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10.5: Perpendicular circles


Assume two circles (Gamma) and (Omega) intersect at two points (X) and (Y). Analogously, (ell') and (m') be the tangent lines at (Y) to (Gamma) and (Omega).

From Exercise 9.6.3, we get that (ell perp m) if and only if (ell' perp m').

We say that the circle (Gamma) is perpendicular to the circle (Omega) (briefly (Gamma perp Omega)) if they intersect and the lines tangent to the circles at one point (and therefore, both points) of intersection are perpendicular.

Similarly, we say that the circle (Gamma) is perpendicular to the line (ell) (briefly (Gamma perp ell)) if (Gamma cap ell e emptyset) and (ell) perpendicular to the tangent lines to (Gamma) at one point (and therefore, both points) of intersection. According to Lemma 5.6.2, it happens only if the line l passes thru the center of (Gamma).

Now we can talk about perpendicular circlines.

Theorem (PageIndex{1})

Assume (Gamma) and (Omega) are distinct circles. Then (Omega perp Gamma) if and only if the circle (Gamma) coincides with its inversion in (Omega).

Proof

Suppose that (Gamma') denotes the inverse of (Gamma).

"Only if" part. Let (O) be the center of (Omega) and (Q) be the center of (Gamma). Let (X) and (Y) denote the points of intersections of (Gamma) and (Omega). According to Lemma 5.6.2, (Omega perp Gamma) if and only if ((OX)) and ((OY)) are tangent to (Gamma).

Note that (Gamma') is also tangent to ((OX)) and ((OY)) and (X) and (Y) respectively. It follows that (X) and (Y) are the foot points of the center of (Gamma') on ((OX)) and ((OY)). Therefore, both (Gamma') and (Gamma) have the center (Q). Finally, (Gamma' = Gamma), since both circles pass thru (X).

"If" part. Assume (Gamma = Gamma').

Since (Gamma e Omega), there is a point (P) that lies on (Gamma), but not on (Omega). Let (P') be the inverse of (P) in (Omega). Since (Gamma = Gamma'), we have that (P' in Gamma). In particular, the half-line ([OP)) intersects (Gamma) at two points. By Exercise 5.6.1, (O) lies outside of (Gamma).

As (Gamma) has points inside and outside of (Omega), the circles (Gamma) and (Omega) intersect. The latter follows from Exercise 3.5.1.

Let (X) be point of their intersection. We need to show that ((OX)) is tangent to (Gamma); that is, (X) is the only intersection point of ((OX)) and (Gamma).

Assume (Z) is another point of intersection. Since (O) is outside of (Gamma), the point (Z) lies on the half-line ([OX)).

Suppose that (Z') denotes the inverse of (Z) in (Omega). Clearly, the three points (Z), (Z'), (X) lie on (Gamma) and ((OX)). The latter contradicts Lemma 5.6.1.

It is convenient to define the inversion in the line (ell) as the reflection across (ell). This way we can talk about inversion in an arbitrary circline.

Corollary (PageIndex{1})

Let (Omega) and (Gamma) be distinct circlines in the inversive plane. Then the inversion in (Omega) sends (Gamma) to itself if and only if (Omega perp Gamma).

Proof

By Thorem (PageIndex{1}), it is sufficient to consider the case when (Omega) or (Gamma) is a line.

Assume (Omega) is a line, so the inversion in (Omega) is a reflection. In this case the statement follows from Corollary 5.4.1.

If (Gamma) is a line, then the statement follows from Theorem 10.3.2.

Corollary (PageIndex{2})

Let (P) and (P') be two distinct points such that (P') is the inverse of (P) in the circle (Omega). Assume that the circline (Gamma) passes thru (P) and (P'). Then (Gamma perp Omega).

Proof

Without loss of generality, we may assume that (P) is inside and (P') is outside (Omega). By Theorem 3.5.1, (Gamma) intersects (Omega). Suppose that A denotes a point of intersection.

Suppose that (Gamma') denotes the inverse of (Gamma). Since (A) is a self-inverse, the points (A, P), and (P') lie on (Gamma'). By Exercise 8.1.1, (Gamma') = (Gamma) and by Theorem (PageIndex{1}), (Gamma perp Omega).

Corollary (PageIndex{3})

Let (P) and (Q) be two distinct points inside the circle (Omega). Then there is a unique circline (Gamma) perpendicular to (Omega) that passes thru (P) and (Q).

Proof

Let (P') be the inverse of the point (P) in the circle (Omega). According to Corollary (PageIndex{2}), the circline is passing thru (P) and (Q) is perpendicular to (Omega) if and only if it passes thru (P').

Note that (P') lies outside of (Omega). Therefore, the points (P), (P'), and (Q) are distinct.

According to Exercise Exercise 8.1.1, there is a unique circline passing thru (P, Q), and (P'). Hence the result.

Exercise (PageIndex{1})

Let (P, Q, P'), and (Q') be points in the Euclidean plane. Assume (P') and (Q') are inverses of (P) and (Q) respectively. Show that the quadrangle (PQP'Q') is inscribed.

Hint

Apply Theorem 10.2.1, Theorem 7.4.5 and Theorem 9.2.1.

Exercise (PageIndex{2})

Let (Omega_1) and (Omega_2) be two perpendicular circles with centers at (O_1) and (O_2) respectively. Show that the inverse of (O_1) in (Omega_2) coincides with the inverse of (O_2) in (Omega_1).

Hint

Suppose that (T) denotes a point of intersection of (Omega_1) and (Omega_2). Let (P) be the foot point of (T) on ((O_1O_2)). Show that ( riangle O_1PT sim riangle O_1TO_2 sim riangle TPO_2). Consider that (P) coincides with the inverses of (O_1) in (Omega_2) and of (O_2) in (Omega_1).

Exercise (PageIndex{3})

Three distinct circles — (Omega_1), (Omega_2) and (Omega_3) , intersect at two points — (A) and (B). Assume that a circle (Gamma) is perpendicular to (Omega_1) and (Omega_2). Show that (Gamma perp Omega_3).

Hint

Since (Gamma perp Omega_1) and (Gamma perp Omega_2), Corollary (PageIndex{1}) implies that the circles (Omega_1) and (Omega_2) are inverted in (Gamma) to themselves. Conclude that the points (A) and (B) are inverse of each other. Since (Omega_3 i A, B), Corollary (PageIndex{2}) implies that (Omega_3 perp Gamma).

Let us consider two new construction tools: the circumtool that constructs a circline thru three given points, and the inversion-tool — a tool that constructs an inverse of a given point in a given circline.

Exercise (PageIndex{4})

Given two circles (Omega_1), (Omega_2) and a point (P) that does not lie on the circles, use only circum-tool and inversion-tool to construct a circline (Gamma) that passes thru (P), and perpendicular to both (Omega_1) and (Omega_2).

Hint

Let (P_1) and (P_2) be the inverse of (P) in (Omega_1) and (Omega_2). Apply Corollary (PageIndex{2}) and Theorem (PageIndex{1}) to show that a circline (Gamma) that pass thru

(P, P_1), and (P_3) is a solution.

Advanced Exercise (PageIndex{5})

Given three disjoint circles (Omega_1), (Omega_2) and (Omega_3), use only circum-tool and inversion-tool to construct a circline (Gamma) that perpendicular to each circle (Omega_1), (Omega_2) and (Omega_3).

Think what to do if two of the circles intersect.

Hint

All circles that perpendicular to (Omega_1) and (Omega_2) pass thru a fixed point (P). Try to construct (P).

If two of the circles intersect, try to apply Corollary 10.6.1.


NCERT Solutions for Class 9 Maths Chapter 10 Exercise 10.5 – Circles

Download NCERT Solutions for Class 9 Maths Chapter 10 Exercise 10.5 – Circles. This Exercise contains 12 questions, for which detailed answers have been provided in this note. In case you are looking at studying the remaining Exercise for Class 9 for Maths NCERT solutions for other Chapters, you can click the link at the end of this Note.

NCERT Solutions for Class 9 Maths Chapter 10 Exercise 10.5 – Circles

NCERT Solutions for Class 9 Maths Chapter 10 Exercise 10.5 – Circles

1. In given figure, A,B and C are three points on a circle with centre O such that BOC = 30° and AOB = 60°. If D is a point on the circle other than the arc ABC, find ADC.

Here, it can be clearly observed that
∠AOC = ∠AOB + ∠BOC
= 60° + 30°
= 90°
We know that,
The angle subtended by an arc at the centre is double the angle subtended by it any point on the remaining part of the circle.
=> ∠ADC = (1/2) ∠ AOC
=> ∠ADC = (1/2) x 90°
=> ∠ADC = 45°

2. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

In ∆ AOB,
AB = OA = OB ( Given )
=> ∆ AOB is an equilateral triangle.
=> ∠ AOB = ∠ OAB = ∠ OBA = 60°
And that implies, ∠ ACB = (1/2) ∠ AOB = 30°
Here, ADBC is a cyclic quadrilateral.
=> ∠ ACB + ∠ ADB = 180°
=> ∠ ADB = 180° – 30°
=> ∠ ADB = 150°
Hence, the angle subtended by this chord at a point on the major arc and the minor arc are 30° and 150° respectively.

3. In given figure, PQR = 100°, where P, Q and R are points on a circle with centre O. Find OPR.

Here, PR is the chord of the circle.
Let, S is any point on the major arc of the circle.
=> PQRS is a cyclic quadrilateral.
∠PQR + ∠ PSR = 180°
=> ∠ PSR = 180° – 100°
=> ∠ PSR = 80°
We know that,
The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
=> ∠ POR = 2 (∠ PSR )
= 2 x 80°
= 160°
In triangle POR,
OP = OR ( Both are Radii )
=> ∠ PRO = ∠ OPR
∠ POR + ∠ PRO + ∠ OPR = 180°
=> 2 ∠ OPR = 180° – 160°
=> 2 ∠ OPR = 20°
=> ∠ OPR = 10°

4. In given figure, ABC = 69°, ACB = 31°, find BDC.

We know that,
Angles subtend by same chord in same segment will be equal.
=> ∠ BAC = ∠ BDC ..(1)
In triangle ABC,
∠ BAC + ∠ ABC + ∠ ACB = 180°
=> ∠ BAC + 69° + 31° = 180°
=> ∠ BAC = 180° – 100°
=> ∠ BAC = 80°
From (1)
∠ BDC = 80°

5. In given figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that BEC = 130° and ECD = 20°. Find BAC.

Since, BED is a straight line
∠ BEC + ∠ CED = 180°
=> ∠ CED = 180° – 130°
=> ∠ CED = 50°
From the triangle EDC,
∠ EDC + ∠ CED + ∠ ECD = 180°
=> ∠ EDC = 180° – 50° – 20°
=> ∠ EDC = 110°
We know that,
∠ EDC = ∠ BDC
=> ∠ BDC = 110°
We know that,
Angles in same segment make by same chord will be equal.
=> ∠ BDC = ∠ BAC
=> ∠ BAC = 110°

6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If DBC = 70°, BAC is 30°, find BCD. Further, if AB = BC, find ECD.

We know that,
Angles in the same segment form by same chord will be equal.
For chord CD,
∠CBD = ∠CAD
=> ∠CAD = 70°
∠BAD = ∠CAD + ∠CAB
= 70° + 30°
= 100°
Since ABCD is a cyclic quadrilateral,
∠BAD + ∠BCD = 180°
∠BCD = 180° – 100°
∠BCD = 80° ..(1)
In the triangle ABC,
AB = BC ( given )
=> ∠BCA = ∠BAC
= 30°
∠BCD = ∠BCA + ∠ACD
=> ∠ ACD = 80° – 30°
=> ∠ ACD = 50°
We know that,
∠ ACD = ∠ ECD
=> ∠ ECD = 50°

7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Here, ABCD is a cyclic quadrilateral intersecting their diagonals AC & BD at center of the circle O.
Consider the chord BD,
=> ∠BAD = (1/2) ∠BOD
=> ∠BAD = (1/2) ( 180°)
=> ∠BAD = 90°
In a cyclic quadrilateral ABCD,
∠BAD + ∠BCD = 180°
=> ∠BCD = 180° – 90°
=> ∠BCD = 90°
Consider the chord AC,
=> ∠ABC = (1/2) ∠AOC
=> ∠ABC = (1/2) ( 180°)
=> ∠ABC = 90°
In a cyclic quadrilateral ABCD,
∠ABC + ∠ADC = 180°
=> ∠ADC = 180° – 90°
=> ∠ADC = 90°
Each interior angle of the cyclic quadrilateral is 90°.
=> Cyclic quadrilateral having diameters of the circle as its diagonals will be a rectangle.

Hence Proved.

8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Here, ABCD is a trapezium with AB || CD.
AD = BC ( Given )
Construction :
Draw perpendiculars from A & B on CD.
=> AM & BN are perpendicular to CD.
From the triangles AMD & BNC,
∠AMD = ∠BNC (Each 90˚)
AM = BN ( Perpendicular distance between parallel lines will be equal )
AD = BC ( given )
=> ∆ AMD & ∆ BNC are congruent under R.H.S. congruency rule.
=> ∠ADC = ∠BCD [By CPCT] ..(1)
( Corresponding angles are equal in congruent triangles )
AB || CD
AD is the transversal.
∠BAD + ∠ADC = 180°
From (1),
∠BAD + ∠BCD = 180° ..(2)
∠ABC + ∠ADC + ∠BAD + ∠BCD = 360°
=> ∠ABC + ∠ADC = 180° ..(3)
Equation (2) & Equation (3) shows that the opposite angles are supplementary.
We can say that,
If the non-parallel sides of a trapezium are equal, then it will be cyclic.
Hence Proved.

9. Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see given figure). Prove that ACP = QCD.

Construction :
Join chords AP & DQ
For the chord AP,
∠ACP = ∠ABP ( Angles by same chord in same segment will be equal ) ..(1)
For the chord DQ,
∠DBQ = ∠QCD ( Angles by same chord in same segment will be equal ) ..(2)
∠ABP = ∠DBQ ( Vertically opposite angles ) ..(3)
From (1), (2) & (3),
We can say that,
∠ACP = ∠ QCD
Hence Proved.

10. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Consider the triangle ABC.
Here AB & AC are diameters of the respective circles.
Those two circles intersect at A & D
D do not lie on BC.
=> ∠ADB = ∠ADC = 90° ( Angle in a semi-circle is 90°)
We know that,
∠BDC = ∠ADB + ∠ADC
= 90° + 90°
= 180°
=> Therefore, BDC is a straight line (D should lie on the line BC)
=> Hence, our assumption was wrong
=> The intersecting point of circles should lie on the third side of the triangle.
Hence Proved.

11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that CAD = CBD.

From the triangle ABC,
∠ABC + ∠BAC + ∠ACB = 180° ( sum of interior angle in a triangle is 180°)
90° + ∠BAC + ∠ACB = 180°
=> ∠BAC + ∠ACB = 90° ..(1)
From the triangle ADC,
∠ADC + ∠DAC + ∠ACD = 180° ( sum of interior angle in a triangle is 180°)
90° + ∠DAC + ∠ACD = 180°
=> ∠DAC + ∠ACD = 90° ..(2)
Adding the equations (1) & (2),
=> ∠BAC + ∠ACB + ∠DAC + ∠ACD = 90° + 90°
= 180°
=> (∠BAC + ∠DAC) + (∠ACB + ∠ACD) = 180°
=> ∠BAD + ∠BCD = 180° ..(3)
We have,
∠ABC + ∠ADC = 180° ..(4)
From (3) & (4),
We can say that, ABCD is a cyclic quadrilateral.

Here it can be clearly see that
∠ CAD = ∠ CBD ( Angles in same segment by same chord will be equal. )

Hence Proved.

12. Prove that a cyclic parallelogram is a rectangle.

Here ABCD is a cyclic parallelogram.
∠ DAB + ∠ DCB = 180° .. (1)
And we know that,
In a parallelogram opposite angles are equal.
=>∠ DAB = ∠ DCB .. (2)
From (1) & (2)
∠ DAB = ∠ DCB = 90°
Parallelogram will be a rectangle if one of its interior angle is 90°.
Hence, cyclic parallelogram is a rectangle.

NCERT Solutions for Class 9 Maths Chapter 10 Exercise 10.5 – Circles, has been designed by the NCERT to test the knowledge of the student on the following topics:-

Download NCERT Solutions for Class 9 Maths Chapter 10 Exercise 10.5 – Circles


&emsp&emsp l must be tangent to circle S .

Explanation of Solution

l is perpendicular to circle s and T S T ¯ is a radius of circle s .

Concept used:

Two triangles having same all three sides equal.

Two triangles having two sides equal and one angle equal.

Two triangles where two angles and one side is equal.

Calculation:

Assume that l in not the tangent to circle S , then it must intersect circle at a point other than T .

Let the point be U . Then the triangle T S U must be an isosceles triangle since, S U = S T .

∠ S T U and ∠ T S U must be both acute because each has to be less than 90 0 in order for ∠ T S U to have a positive degree measure. However, if ∠ S T U is acute, the line U T will not be perpendicular to radius S T , which is contradictory to the given.

The diagram of the assumption is:

The assumption is false that it is not true that l in not the tangent to circle S .


NCERT Solutions For Class 9 Maths Chapter 10: Circles

The NCERT Solutions for Class 9 Maths Chapter 10 involves exercise 10.1, exercise 10.2, exercise 10.3, exercise 10.4, exercise 10.5, and exercise 10.6. Before we get into the detailed NCERT Solutions for Class 9 Maths Chapter 10, let’s look at the topics included in this chapter:

ExerciseTopics
10.1Introduction
10.2Circles and its related terms: A Review
10.3Angle Subtended by a chord at a point
10.4Perpendicular from the Centre to a Chord
10.5Circle through three points
10.6Equal Chords and their distances from the Centre
10.7Angle Subtended by an Arc of a Circle
10.8Cyclic Quadrilaterals
10.9Summary

CBSE Class 9 Maths Chapter 10: Chapter Summary

In this chapter, students will learn about circles, its related terms, and the properties of a circle. A circle is a collection of points on a plane, which are equidistant from a fixed point in the plane. A circle is the locus of points equidistant from a given point. The fixed point is called the center while the fixed distance is called the radius.

A circle divides the plane into three parts: inside the circle, on the circle, and the area outside the circle. In this chapter, students will learn some terms related to circles like the center of a circle, radius, diameter, arcs, and its types – major arc and minor arc, semicircle, chords, cyclic quadrilaterals, segments, and its two types – major segment and minor segment, etc.

How To Prepare For CBSE Class 9 Maths Circles?

The Class 9 Maths Circles notes (solutions) are provided here in a simplified manner which is easy to follow and understand. Once your concepts are clear, solving the exercise questions becomes quite easy. Make the best use of the solutions provided here as you solve the exercise questions.

To boost up your performance in the exam, you can solve free Class 9 Maths Circle questions at Embibe. Also, students can take CBSE Class 9 Circles Chapter-wise mock test to test their performance. Finish the chapter in the most effective manner so that you can solve any question asked.

NCERT SOLUTIONS FOR CLASS 9 MATHS CHAPTER 9NCERT SOLUTIONS FOR CLASS 9 MATHSNCERT SOLUTIONS FOR CLASS 9 MATHS CHAPTER 11

CBSE Class 9 Maths Chapter 10 Important Questions

Some of the important questions from this chapter are as under:

  1. Bisectors of angles A, B, and C of a triangle ABC intersect its circumcircle at D, E, and F respectively. Prove that the angles of the triangle DEF are 90° – (½)A, 90° – (½)B and 90° – (½)C.
  2. In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
  3. Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
  4. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lies on the third side.
  5. Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its center. If the distance between AB and CD is 6, find the radius of the circle.
  6. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
  7. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.
  8. A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed, and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone.
  9. Prove that if chords of congruent circles subtend equal angles at their centers, then the chords are equal.
  10. Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the center.
  11. AC and BD are chords of a circle which bisect each other. Prove that
    (i) AC and BD are diameters
    (ii) ABCD is a rectangle.
  12. Prove that a cyclic parallelogram is a rectangle.

FAQs on NCERT Solutions for Class 9 Maths Chapter 10

Here are the frequently asked questions (FAQs) on NCERT Solutions for Class 9 Maths Chapter 10:

Q1. What can I learn from NCERT Class 9 Maths Chapter 10?

A. In this chapter, you will learn about circles, its related terms, and the properties of a circle. Moreover, you will also learn some terms related to circles like the center of a circle, radius, diameter, arcs, and its types – major arc and minor arc, semicircle, chords, cyclic quadrilaterals, segments, and its two types – major segment and minor segment, etc.

Q2. Which topics are covered in NCERT Class 9 Maths Chapter 10?

A. The topics involved in the NCERT Class 9 Maths Chapter 10 are as follows:

  • 10.1: Introduction
  • 10.2: Circles and its related terms: A Review
  • 10.3: Angle Subtended by a chord at a point
  • 10.4: Perpendicular from the Centre to a Chord
  • 10.5: Circle through three points
  • 10.6: Equal Chords and their distances from the Centre
  • 10.7: Angle Subtended by an Arc of a Circle
  • 10.8: Cyclic Quadrilaterals
  • 10.9: Summary

Q3. How many exercises are present in NCERT Class 9 Maths Chapter 10?

A. The NCERT Class 9 Maths Chapter 10 is divided into 9 sections and 6 exercises. The NCERT Solutions for Class 9 Maths Chapter 10 involves exercise 10.1, exercise 10.2, exercise 10.3, exercise 10.4, exercise 10.5, and exercise 10.6.

Q4. What is the difference between a chord and a diameter?

A. A chord is described as a line segment drawn between any two points in a circle. However, a diameter is drawn at the middle of a circle. It cuts the circle into two halves.

Q5. Are the NCERT Solutions for Class 9 Maths Chapter 10 easy to understand?

A.The NCERT Solutions by Embibe will help you in understanding the chapter Circle better. The NCERT Solutions for Class 9 Maths Chapter 10 have been explained in a step by step manner and detailed manner. So that any class 9 student can understand NCERT Solutions for Class 9 Chapter 10.

Q6. What are the benefits of using NCERT Solutions for Class 9 Maths Chapter 10?

A. The concepts are explained in simple language in the NCERT Solutions for Class 9 Chapter 10 by Embibe. To boost up your performance in the exam, you can solve free Class 9 Maths Circle questions at Embibe. Also, students can take CBSE Class 9 Circles Chapter-wise mock test to test their performance. These are some of the benefits of using NCERT Solutions for Class 9 Maths Chapter 10.

We hope this article on NCERT Solutions for Class 9 Maths Chapter 10 – Circles helps you. If you have any questions, feel free to ask in the comment section below. We will get back to you at the earliest.


Properties of Perpendicular Lines

We have already seen how the perpendicular lines look like. If there is an "L" shape in a figure, the corresponding angle at the vertex is a right angle. Perpendicular lines always intersect each other, however, all intersecting lines are not always perpendicular to each other. The two main properties of perpendicular lines are:

  • Perpendicular lines always meet or intersect each other.
  • The angle between any two perpendicular lines is always equal to 90 °

10.5: Perpendicular circles

Tangents

A tangent is a line that just skims the surface of a circle. It hits the circle at one point only.

There are two main theorems that deal with tangents. The first one is as follows:

A tangent line of a circle will always be perpendicular to the radius of that circle. It will always form a right angle (90°) with the radius.


Questions that deal with this theorem usually go hand in hand with the Pythagorean theorem. That’s because you can only use this theorem if you have a right triangle. The Pythagorean theorem is: ( + = ) where “c” is always the hypotenuse.

Is () a tangent?

If () is a tangent, then that should be a right triangle which means the Pythagorean theorem will work.

(egin
+ = \
<3^2>+ <12^2>= <15^2>\
9 + 144 = 225\
153 e 225
end)
It doesn’t work, so () is not a tangent!

() is a tangent. Find x.

There is another theorem that deals with tangents as well.

If two tangents to the same circle share a point outside of the circle, then the two tangents are congruent.

If () and () are tangents, then (overline cong overline )

(overline ) and () are tangents. Find x.

Since both lines are tangents and share the point B, then they are equal.

(overline ) and () are tangents. Find x.

These lines are equal as well.

All lines are tangents. Find the perimeter of the polygon.

To find perimeter, add up all the numbers.

8 + 8 + 3.9 + 3.9 + 8 + 8 + 3.9 + 3.9 = 47.6

Let’s try one last example.

All lines are tangents. Find the perimeter of the polygon.

This one is a little bit tougher. We have to figure it out piece by piece.


Circles ML Aggarwal Solutions ICSE Class-10 Maths Chapter-15

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How to Solve Circles Problems/Questions / Exercise of ICSE Class-10 Mathematics

Before viewing Answer of Chapter-15 Circles of ML Aggarwal Solutions. Read the Chapter Carefully and then solve all example given in your text book.

For more practice on Circles related problems /Questions / Exercise try to solve Circles exercise of other famous publications also such as Goyal Brothers Prakshan (RS Aggarwal ICSE) / Concise Selina Publications ICSE Mathematics. Get the formula of Circles for ICSE Class 10 Maths to understand the topic more clearly in effective way.

Exercise 15.1 , Circle ML Aggarwal Solutions

Question 1

Using the given information, find the value of x in each of the following figures :

Answer 1

(i) ∠ADB and ∠ACB are in the same segment.
∠ADB = ∠ACB = 50°
Now in ∆ADB,
∠DAB + X + ∠ADB = 180°



Question 2

If O is the centre of the circle, find the value of x in each of the following figures (using the given information):



Answer 2

(Angles in the same segment of a circle)
But ∠ADB = x°



Question 3

(a) In the figure (i) given below, AD || BC. If ∠ACB = 35°. Find the measurement of ∠DBC.
(b) In the figure (ii) given below, it is given that O is the centre of the circle and ∠AOC = 130°. Find ∠ABC


Answer 3

(a) Construction: Join AB
∠A = ∠C = 35° [∵ Alt angles]

Question 4

(a) In the figure (i) given below, calculate the values of x and y.
(b) In the figure (ii) given below, O is the centre of the circle. Calculate the values of x and y.


Answer 4

(a) ABCD is a cyclic quadrilateral

Question 5

(a) In the figure (i) given below, M, A, B, N are points on a circle having centre O. AN and MB cut at Y. If ∠NYB = 50° and ∠YNB = 20°, find ∠MAN and the reflex angle MON.

(b) In the figure (ii) given below, O is the centre of the circle. If ∠AOB = 140° and ∠OAC = 50°, find
(i) ∠ACB
(ii) ∠OBC
(iii) ∠OAB
(iv) ∠CBA


Answer 5

(a) ∠NYB = 50°, ∠YNB = 20°.

Question 6

(a) In the figure (i) given below, O is the centre of the circle and ∠PBA = 42°. Calculate the value of ∠PQB
(b) In the figure (ii) given below, AB is a diameter of the circle whose centre is O. Given that ∠ECD = ∠EDC = 32°, calculate
(i) ∠CEF
(ii) ∠COF.

Answer 6

(a) In ∆APB,
∠APB = 90° (Angle in a semi-circle)
But ∠A + ∠APB + ∠ABP = 180° (Angles of a triangle)
∠A + 90° + 42°= 180°
∠A + 132° = 180°
⇒ ∠A = 180° – 132° = 48°
But ∠A = ∠PQB

Question 7

(a) In the figure (i) given below, AB is a diameter of the circle APBR. APQ and RBQ are straight lines, ∠A = 35°, ∠Q = 25°. Find (i) ∠PRB (ii) ∠PBR (iii) ∠BPR.
(b) In the figure (ii) given below, it is given that ∠ABC = 40° and AD is a diameter of the circle. Calculate ∠DAC.

Answer 7


(a) (i) ∠PRB = ∠BAP
(Angles in the same segment of the circle)
∴ ∠PRB = 35° (∵ ∠BAP = 35° given)
(ii) In ∆PRQ,

Question 8

(a) In the figure given below, P and Q are centers of two circles intersecting at B and C. ACD is a straight. line. Calculate the numerical value of x.

(b) In the figure given below, O is the circumcenter of triangle ABC in which AC = BC. Given that ∠ACB = 56°, calculate
(i)∠CAB
(ii)∠OAC


Answer 8

Given that
(a) Arc AB subtends ∠APB at the centre
and ∠ACB at the remaining part of the circle

Question 9

(a) In the figure (i) given below, chord ED is parallel to the diameter AC of the circle. Given ∠CBE = 65°, calculate ∠DEC.

(b) In the figure (ii) given below, C is a point on the minor arc AB of the circle with centre O. Given ∠ACB = p°, ∠AOB = q°, express q in terms of p. Calculate p if OACB is a parallelogram.

Answer 9

(a) ∠CBE = ∠CAE
(Angle in the same segment of a circle)
⇒ ∠CAE = 65°
∠AEC = 90° (Angle in a semi circle)
Now in ∆AEC
∠AEC + ∠CAE + ∠ACE = 180° (Angle of a triangle)
⇒ 90°+ 65° +∠ACE = 180°
so ⇒ 155° + ∠ACE = 180°
hence ⇒ ∠ACE = 180° – 155° – 25°
∵AC || ED (given)
∴∠ACE = ∠DEC (alternate angles)
∴∠DEC = 25°

Question 10

(a) In the figure (i) given below, straight lines AB and CD pass through the centre O of a circle. If ∠OCE = 40° and ∠AOD = 75°, find the number of degrees in :
(i) ∠CDE
(ii) ∠OBE.
(b) In the figure (ii) given below, I is the incentre of ∆ABC. AI produced meets the circumcircle of ∆ABC at D. Given that ∠ABC = 55° and ∠ACB = 65°, calculate
(i) ∠BCD
(ii) ∠CBD
(iii) ∠DCI
(iv) ∠BIC.


Answer 10

(a) (i) ∠CED = 90° (Angle in semi-circle)
In ∆CED
∠CED + ∠CDE + ∠DCE = 180°
so ⇒ 90° +∠CDE + 40° = 180°
therefore ⇒ 130° + ∠CDE = 180°
hence ⇒ ∠CDE = 180° – 130° = 50°


Question 11

O is the circumcenter of the triangle ABC and D is mid-point of the base BC. Prove that ∠BOD = ∠A.

Answer 11

In the given figure, O is the centre of circumcenter of ∆ABC.
D is mid-point of BC. BO, CO and OD are joined.

Question 12

In the adjoining figure, AB and CD are equal chords. AD and BC intersect at E. Prove that AE = CE and BE = DE.

Answer 12

In the given figure, AB and CD are two equal chords
AD and BC intersect each other at E.
To prove : AE = CE and BE = DE
Proof:
In ∆AEB and ∆CED
AB = CD (given)
∠A = ∠C (angles in the same segment)
∠B = ∠D (angles in the same segment)
∴ ∆AEB ≅ ∆CED (ASA axiom)
∴ AE = CE and BE = DE (c.p.c.t.)

Question 13

(a) In the figure (i) given below, AB is a diameter of a circle with centre O. AC and BD are perpendiculars on a line PQ. BD meets the circle at E. Prove that AC = ED.

(b) In the figure (ii) given below, O is the centre of a circle. Chord CD is parallel to the diameter AB. If ∠ABC = 25°, calculate ∠CED.


Answer 13

(a) Given: AB is the diameter of a circle with centre O.
AC and BD are perpendiculars on a line PQ,
such that BD meets the circle at E.


Question 14

In the adjoining figure, O is the centre of the given circle and OABC is a parallelogram. BC is produced to meet the circle at D.

Prove that ∠ABC = 2 ∠OAD.

Answer 14

Given: In the figure,
OABC is a || gm and O is the centre of the circle.
BC is produced to meet the circle at D.
To Prove : ∠ABC = 2∠OAD.
Construction: Join AD.

Question 15

(a) In the figure (i) given below, P is the point of intersection of the chords BC and AQ such that AB = AP. Prove that CP = CQ.


(b) In the figure (i) given below, AB = AC = CD, ∠ADC = 38°. Calculate :
(i) ∠ABC (ii) ∠BEC.


Answer 15

(a) Given: Two chords AQ and BC intersect each other at P
inside the circle. AB and CQ are joined and AB = AP.
To Prove : CP = CQ
Construction : Join AC.
Proof: In ∆ABP and ∆CQP
∴ ∠B = ∠Q

Question 16

(a) In the figure (i) given below, CP bisects ∠ACB. Prove that DP bisects ∠ADB.
(b) In the figure (ii) given below, BD bisects ∠ABC. Prove that


Answer 16

(a)Given: In the figure, CP is the bisector of

∠ACB meeting the circle at P.
PD is joined


Question 17

(a) In the figure (ii) given below, chords AB and CD of a circle intersect at E.
(i) Prove that triangles ADE and CBE are similar.
(ii) Given DC =12 cm, DE = 4 cm and AE = 16 cm, calculate the length of BE.

(b) In the figure (ii) given below, AB and CD are two intersecting chords of a circle. Name two triangles which are similar. Hence, calculate CP given that AP = 6cm, PB = 4 cm, and CD = 14 cm (PC > PD).


Answer 17

(a) Given: Two chords AB and CD intersect each other

at E inside the circle.



Question 18

In the adjoining figure, AE and BC intersect each other at point D. If ∠CDE = 90°, AB = 5 cm, BD = 4 cm and CD = 9 cm, find DE. (2008)


Answer 18

In the figure, AE and BC intersect each other at D.
AB is joined.

Question 19

(a) In the figure (i) given below, PR is a diameter of the circle, PQ = 7 cm, QR = 6 cm and RS = 2 cm. Calculate the perimeter of the cyclic quadrilateral PQRS.
(b) In the figure (ii) given below, the diagonals of a cyclic quadrilateral ABCD intersect in P and the area of the triangle APB is 24 cm². If AB = 8 cm and CD = 5 cm, calculate the area of ∆DPC.


Answer 19

(a) PR is the diameter of the circle
PQ = 7 cm, QR = 6 cm, RS = 2 cm.


Question 20

(a) In the figure (i) given below, QPX is the bisector of ∠YXZ of the triangle XYZ. Prove that XY : XQ = XP : XZ,
(b) In the figure (ii) given below, chords BA and DC of a circle meet at P. Prove that:
(i) ∠PAD = ∠PCB
(ii) PA. PB = PC . PD.


Answer 20

(a) Given: ∆XYZ is inscribed in a circle.
Bisector of ∠YXZ meets the circle at Q.
QY is joined.
To Prove : XY : XQ = XP : XZ


Exercise 15.2 Circles ML Aggarwal Solutions ICSE

Question 1

If O is the centre of the circle, find the value of x in each of the following figures (using the given information):


Answer 1

From the figure
(i) ABCD is a cyclic quadrilateral


Question 2

(a) In the figure (i) given below, O is the centre of the circle. If ∠AOC = 150°, find (i) ∠ABC (ii) ∠ADC.

(b) In the figure (i) given below, AC is a diameter of the given circle and ∠BCD = 75°. Calculate the size of (i) ∠ABC (ii) ∠EAF.


Answer 2

(a) Given, ∠AOC = 150° and AD = CD
We know that an angle subtends by an arc of a circle
at the centre is twice the angle subtended by the same arc
at any point on the remaining part of the circle.

Question 3

(a) In the figure, (i) given below, if ∠DBC = 58° and BD is a diameter of the circle, calculate:
(i) ∠BDC (ii) ∠BEC (iii) ∠BAC

(b) In the figure (if) given below, AB is parallel to DC, ∠BCE = 80° and ∠BAC = 25°. Find:
(i) ∠CAD (ii) ∠CBD (iii) ∠ADC (2008)


Answer 3

(a) ∠DBC = 58°
BD is diameter
∠DCB = 90° (Angle in semi circle)
(i) In ∆BDC
∠BDC + ∠DCB + ∠CBD = 180°
∠BDC = 180°- 90° – 58° = 32°


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Question 4

(a) In the figure given below, ABCD is a cyclic quadrilateral. If ∠ADC = 80° and ∠ACD = 52°, find the values of ∠ABC and ∠CBD.

(b) In the figure given below, O is the centre of the circle. ∠AOE =150°, ∠DAO = 51°. Calculate the sizes of ∠BEC and ∠EBC.


Answer 4

(a) In the given figure, ABCD is a cyclic quadrilateral
∠ADC = 80° and ∠ACD = 52°
To find the measure of ∠ABC and ∠CBD


Question 5

(a) In the figure (i) given below, ABCD is a parallelogram. A circle passes through A and D and cuts AB at E and DC at F. Given that ∠BEF = 80°, find ∠ABC.
(b) In the figure (ii) given below, ABCD is a cyclic trapezium in which AD is parallel to BC and ∠B = 70°, find:
(i)∠BAD (ii) DBCD.


Answer 5


(a) ADFE is a cyclic quadrilateral
Ext. ∠FEB = ∠ADF
⇒ ∠ADF = 80°
ABCD is a parallelogram
∠B = ∠D = ∠ADF = 80°
or ∠ABC = 80°
(b)In trapezium ABCD, AD || BC
(i) ∠B + ∠A = 180°
⇒ 70° + ∠A = 180°
⇒ ∠A = 180° – 70° = 110°
∠BAD = 110°
(ii) ABCD is a cyclic quadrilateral
∠A + ∠C = 180°
⇒ 110° + ∠C = 180°
⇒ ∠C = 180° – 110° = 70°
∠BCD = 70°

Question 6

(a) In the figure given below, O is the centre of the circle. If ∠BAD = 30°, find the values of p, q and r.

(b) In the figure given below, two circles intersect at points P and Q. If ∠A = 80° and ∠D = 84°, calculate
(i) ∠QBC (ii) ∠BCP


Answer 6

(a) (i) ABCD is a cyclic quadrilateral

Question 7

(a) In the figure given below, PQ is a diameter. Chord SR is parallel to PQ.Given ∠PQR = 58°, calculate (i) ∠RPQ (ii) ∠STP
(T is a point on the minor arc SP)

(b) In the figure given below, if ∠ACE = 43° and ∠CAF = 62°, find the values of a, b and c (2007)


Answer 7

(a) In ∆PQR,
∠PRQ = 90° (Angle in a semi circle) and ∠PQR = 58°
∠RPQ = 90° – ∠PQR = 90° – 58° = 32°
SR || PQ (given)

Question 8

(a) In the figure (i) given below, AB is a diameter of the circle. If ∠ADC = 120°, find ∠CAB.
(b) In the figure (ii) given below, sides AB and DC of a cyclic quadrilateral ABCD are produced to meet at E, the sides AD and BC are produced to meet at F. If x : y : z = 3 : 4 : 5, find the values of x, y and z.


Answer 8

(a) Construction: Join BC, and AC then
ABCD is a cyclic quadrilateral.


Question 9

(a) In the figure (i) given below, ABCD is a quadrilateral inscribed in a circle with centre O. CD is produced to E. If ∠ADE = 70° and ∠OBA = 45°, calculate
(i) ∠OCA (ii) ∠BAC
(b) In figure (ii) given below, ABF is a straight line and BE || DC. If ∠DAB = 92° and ∠EBF = 20°, find :
(i) ∠BCD (ii) ∠ADC.


Answer 9

(a) ABCD is a cyclic quadrilateral


Question 10

(a) In the figure (ii) given below, PQRS is a cyclic quadrilateral in which PQ = QR and RS is produced to T. If ∠QPR = 52°, calculate ∠PST.

(b) In the figure (ii) given below, O is the centre of the circle. If ∠OAD = 50°, find the values of x and y.


Answer 10

(a) PQRS is a cyclic quadrilateral in which
PQ = QR

Question 11

(a) In the figure (i) given below, O is the centre of the circle. If ∠COD = 40° and ∠CBE = 100°, then find :
(i) ∠ADC
(ii) ∠DAC
(iii) ∠ODA
(iv) ∠OCA.
(b) In the figure (ii) given below, O is the centre of the circle. If ∠BAD = 75° and BC = CD, find :
(i) ∠BOD
(ii) ∠BCD
(iii) ∠BOC
(iv) ∠OBD (2009)


Answer 11

(a) (i) ∴ ABCD is a cyclic quadrilateral.
∴ Ext. ∠CBE = ∠ADC
⇒ ∠ADC = 100°
(ii) Arc CD subtends ∠COD at the centre
and ∠CAD at the remaining part of the circle
∴ ∠COD = 2 ∠CAD


Question 12

In the adjoining figure, O is the centre and AOE is the diameter of the semicircle ABCDE. If AB = BC and ∠AEC = 50°, find :
(i) ∠CBE
(ii) ∠CDE
(iii) ∠AOB.
Prove that OB is parallel to EC.


Answer 12

O is the centre of the semi-circle ABCDE
and AOE is the diameter. AB = BC, ∠AEC = 50°

Question 13

(a) In the figure (i) given below, ED and BC are two parallel chords of the circle and ABE, ACD are two st. lines. Prove that AED is an isosceles triangle.

(b) In the figure (ii) given below, SP is the bisector of ∠RPT and PQRS is a cyclic quadrilateral. Prove that SQ = RS.


Answer 13

ABE and ACD are straight lines.
To Prove: ∆AED is an isosceles triangle.
Proof: BCDE is a cyclic quadrilateral.
Ext. ∠ABC = ∠D …(i)
But BC || ED (given)


Question 14

In the adjoining figure, ABC is an isosceles triangle in which AB = AC and circle passing through B and C intersects sides AB and AC at points D and E. Prove that DE || BC.


Answer 14

In the given figure,
∆ABC is an isosceles triangle in which AB = AC.
A circle passing through B and C intersects
sides AB and AC at D and E.
To prove: DE || BC
Construction : Join DE.
∵ AB = AC
∠B = ∠C (angles opposite to equal sides)
But BCED is a cyclic quadrilateral
Ext. ∠ADE = ∠C
= ∠B (∵ ∠C = ∠B)
But these are corresponding angles
DE || BC
Hence proved.

Question 15

(a) Prove that a cyclic parallelogram is a rectangle.
(b) Prove that a cyclic rhombus is a square.

Answer 15


(a) ABCD is a cyclic parallelogram.

To prove: ABCD is a rectangle
Proof: ABCD is a parallelogram
∠A = ∠C and ∠B = ∠D

Question 16

In the adjoining figure, chords AB and CD of the circle are produced to meet at O. Prove that triangles ODB and OAC are similar. Given that CD = 2 cm, DO = 6 cm and BO = 3 cm, area of quad. CABD

Answer 16

In the given figure, AB and CD are chords of a circle.
They are produced to meet at O.
To prove : (i) ∆ODB

∆OAC
If CD = 2 cm, DO = 6 cm, and BO = 3 cm
To find : AB and also area of the

Construction : Join AC and BD

Exercise 15.3 Circles ML Aggarwal Solutions

Question 1

Find the length of the tangent drawn to a circle of radius 3 cm, from a point distant 5 cm from the centre.

Answer 1

In a circle with centre O and radius 3 cm
and P is at a distance of 5 cm.

Question 2

A point P is at a distance 13 cm from the centre C of a circle and PT is a tangent to the given circle. If PT = 12 cm, find the radius of the circle.

Answer 2

CT is the radius
CP = 13 cm and tangent PT = 12 cm

Question 3

The tangent to a circle of radius 6 cm from an external point P, is of length 8 cm. Calculate the distance of P from the nearest point of the circle.


Answer 3

Radius of the circle = 6 cm
and length of tangent = 8 cm
Let OP be the distance
i.e. OA = 6 cm, AP = 8 cm .
OA is the radius

Question 4

Two concentric circles are of the radii 13 cm and 5 cm. Find the length of the chord of the outer circle which touches the inner circle.

Answer 4

Two concentric circles with centre O
OP and OB are the radii of the circles respectively, then
OP = 5 cm, OB = 13 cm.

Question 5

Two circles of radii 5 cm and 2.8 cm touch each other. Find the distance between their centers if they touch :
(i) externally
(ii) internally.

Answer 5

Radii of the circles are 5 cm and 2.8 cm.
i.e. OP = 5 cm and CP = 2.8 cm.

(i) When the circles touch externally,
then the distance between their centers = OC = 5 + 2.8 = 7.8 cm.
(ii) When the circles touch internally,
then the distance between their centers = OC = 5.0 – 2.8 = 2.2 cm

Question 6

(a) In figure (i) given below, triangle ABC is circumscribed, find x.
(b) In figure (ii) given below, quadrilateral ABCD is circumscribed, find x.


Answer 6

(a) From A, AP and AQ are the tangents to the circle
∴ AQ = AP = 4cm

Question 7

(a) In figure (i) given below, quadrilateral ABCD is circumscribed find the perimeter of quadrilateral ABCD.
(b) In figure (ii) given below, quadrilateral ABCD is circumscribed and AD ⊥ DC find x if radius of incircle is 10 cm.

Answer 7


(a) From A, AP and AS are the tangents to the circle
∴AS = AP = 6
From B, BP and BQ are the tangents
∴BQ = BP = 5
From C, CQ and CR are the tangents

Question 8

(a) In the figure (i) given below, O is the centre of the circle and AB is a tangent at B. If AB = 15 cm and AC = 7.5 cm, find the radius of the circle.
(b) In the figure (ii) given below, from an external point P, tangents PA and PB are drawn to a circle. CE is a tangent to the circle at D. If AP = 15 cm, find the perimeter of the triangle PEC.


Answer 8


(i) Join OB
∠OBA = 90°
(Radius through the point of contact is
perpendicular to the tangent)


Question 9

(a) If a, b, c are the sides of a right triangle where c is the hypotenuse, prove that the radius r of the circle which touches the sides of the triangle is given by

(b) In the given figure, PB is a tangent to a circle with centre O at B. AB is a chord of length 24 cm at a distance of 5 cm from the centre. If the length of the tangent is 20 cm, find the length of OP.


Answer 9


(a) Let the circle touch the sides BC, CA and AB
of the right triangle ABC at points D, E and F respectively,
where BC = a, CA = b
and AB = c (as showing in the given figure).



Question 10

Three circles of radii 2 cm, 3 cm and 4 cm touch each other externally. Find the perimeter of the triangle obtained on joining the centers of these circles.

Answer 10

Three circles with centers A, B and C touch each other externally
at P, Q and R respectively and the radii of these circles are
2 cm, 3 cm and 4 cm.

Question 11

(a) In the figure (i) given below, the sides of the quadrilateral touch the circle. Prove that AB + CD = BC + DA.
(b) In the figure (ii) given below, ABC is triangle with AB = 10cm, BC = 8cm and AC = 6cm (not drawn to scale). Three circles are drawn touching each other with vertices A, B and C as their centers. Find the radii of the three circles.


Answer 11


(a) Given: Sides of quadrilateral ABCD touch the circle at
P, Q, R and S respectively.

Question 12

(a) ln the figure (i) PQ = 24 cm, QR = 7 cm and ∠PQR = 90°. Find the radius of the inscribed circle ∆PQR

(b) In the figure (ii) given below, two concentric circles with centre O are of radii 5 cm and 3 cm. From an external point P, tangents PA and PB are drawn to these circles. If AP = 12cm, find BP.


Answer 12


(a) In the figure, a circle is inscribed in the triangle PQR
which touches the sides. O is centre of the circle.
PQ = 24cm, QR = 7 cm ∠PQR = 90°
OM is joined.


Question 13

(a) In the figure (i) given below, AB = 8 cm and M is mid-point of AB. Semi-circles are drawn on AB, AM and MB as diameters. A circle with centre C touches all three semi-circles as shown, find its radius.

(b) In the figure (ii) given below, equal circles with centres O and O’ touch each other at X. OO’ is produced to meet a circle O’ at A. AC is tangent to the circle whose centre is O. O’D is perpendicular to AC. Find the value of :
(i)
(ii)


Answer 13

(a) Let x be the radius of the circle
with centre C and radii of each equal

Question 14

The length of the direct common tangent to two circles of radii 12 cm and 4 cm is 15 cm. Calculate the distance between their centers.


Answer 14

Let R and r be the radii of the circles
with centre A and B respectively
Let TT’ be their common tangent.

Hence distance between their centers = 17 cm Ans.

Question 15

Calculate the length of a direct common tangent to two circles of radii 3 cm and 8 cm with their centers 13 cm apart.


Answer 15

Let A and B be the centers of the circles
whose radii are 8 cm and 3 cm and
let TT’ length of their common tangent and AB = 13 cm.

Question 16

In the given figure, AC is a transverse common tangent to two circles with centres P and Q and of radii 6 cm and 3 cm respectively. Given that AB = 8 cm, calculate PQ.


Answer 16


AC is a transverse common tangent to the two circles
with centre P and Q and of radii 6 cm and 3 cm respectively
AB = 8 cm. Join AP and CQ.

Question 17

Two circles with centers A, B are of radii 6 cm and 3 cm respectively. If AB = 15 cm, find the length of a transverse common tangent to these circles.


Answer 17

AB = 15 cm.
Radius of the circle with centre A = 6 cm
and radius of second circle with radius B = 3 cm.

Question 18

(a) In the figure (i) given below, PA and PB are tangents at a points A and B respectively of a circle with centre O. Q and R are points on the circle. If ∠APB = 70°, find (i) ∠AOB (ii) ∠AQB (iii) ∠ARB
(b) In the figure (ii) given below, two circles touch internally at P from an external point Q on the common tangent at P, two tangents QA and QB are drawn to the two circles. Prove that QA = QB.


Answer 18

(a) To find : (i) ∠AOB, (ii) ∠AQB, (iii) ∠ARB
Given: PA and PB are tangents at the points A and B respectively
of a circle with centre O and OA and OB are radii on it.
∠APB = 70°
Construction: Join AB

Question 19


In the given figure, AD is a diameter of a circle with centre O and AB is tangent at A. C is a point on the circle such that DC produced intersects the tangent of B. If ∠ABC = 50°, find ∠AOC.


Answer 19


Given AB is tangent to the circle at A and OA is radius, OA ⊥ AB
In ∆ABD

Question 20

In the given figure, tangents PQ and PR are drawn from an external point P to a circle such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQ, Find ∠RQS


Answer 20

In the given figure,
PQ and PR are tangents to the circle with centre O drawn from P
∠RPQ = 30°
Chord RS || PQ is drawn
To find ∠RQS
∴ PQ = PR (tangents to the circle)

Question 21

(a) In the figure (i) given below, PQ is a tangent to the circle at A, DB is a diameter, ∠ADB = 30° and ∠CBD = 60°, calculate (i) ∠QAB (ii) ∠PAD (iii) ∠CDB.

(b) In the figure (ii) given below, ABCD is a cyclic quadrilateral. The tangent to the circle at B meets DC produced at F. If ∠EAB = 85° and ∠BFC = 50°, find ∠CAB.


Answer 21


(a) PQ is tangent and AD is chord
(i) ∴ ∠QAB = ∠BDA = 30°
(Angles in the alternate segment)
(ii) In ∆ADB,
∠DAB = 90° (Angle in a semi-circle)


⇒ ∠CAB = 35°

Question 22

(a) In the figure (i) given below, O is the centre of the circle and SP is a tangent. If ∠SRT = 65°, find the value of x, y and z. (2015)
(b) In the figure (ii) given below, O is the centre of the circle. PS and PT are tangents and ∠SPT = 84°. Calculate the sizes of the angles TOS and TQS.


Answer 22


Consider the following figure:
TS ⊥ SP,
∠TSR = ∠OSP = 90°
In ∆TSR,
∠TSR + ∠TRS + ∠RTS = 180°

Question 23

In the adjoining figure, O is the centre of the circle. Tangents to the circle at A and B meet at C. If ∠ACO = 30°, find
(i) ∠BCO (ii) ∠AOR (iii) ∠APB


Answer 23


(i) ∠BCO = ∠ACO = 30°
(∵ C is the intersecting point of tangent AC and BC)
(ii) ∠OAC = ∠OBC = 90°
∵∠AOC = ∠BOC = 180° – (90° + 30°) = 60°
(∵ sum of the three angles a ∆ is 180°)

Question 24


(a) In the figure (i) given below, O is the centre of the circle. The tangent at B and D meet at P. If AB is parallel to CD and ∠ ABC = 55°. find: (i)∠BOD (ii) ∠BPD
(b) In the figure (ii) given below. O is the centre of the circle. AB is a diameter, TPT’ is a tangent to the circle at P. If ∠BPT’ = 30°, calculate : (i)∠APT (ii) ∠BOP.


Answer 24


(a) AB || CD
(i) ∠ABC = ∠BCD (Alternate angles)
⇒ ∠BCD = 55°

Question 25

In the adjoining figure, ABCD is a cyclic quadrilateral.

The line PQ is the tangent to the circle at A. If ∠CAQ : ∠CAP = 1 : 2, AB bisects ∠CAQ and AD bisects ∠CAP, then find the measure of the angles of the cyclic quadrilateral. Also prove that BD is a diameter of the circle.

Answer 25


ABCD is a cyclic quadrilateral.
PAQ is the tangent to the circle at A.
∠CAD : ∠CAP = 1 : 2.
AB and AD are the bisectors of ∠CAQ and ∠CAP respectively

Question 26

In a triangle ABC, the incircle (centre O) touches BC, CA and AB at P, Q and R respectively. Calculate (i) ∠QOR (ii) ∠QPR given that ∠A = 60°.

Answer 26


OQ and OR are the radii and AC and AB are tangents.
OQ ⊥ AC and OR ⊥ AB
Now in the quad. AROQ

Question 27

(a) In the figure (0 given below, AB is a diameter. The tangent at C meets AB produced at Q, ∠CAB = 34°. Find
(i) ∠CBA (ii) ∠CQA (2006)

(b) In the figure (ii) given below, AP and BP are tangents to the circle with centre O. Given ∠APB = 60°, calculate.
(i) ∠AOB (ii) ∠OAB (iii) ∠ACB.


Answer 27

(a) AB is the diameter.

Question 28


(a) In the figure (i) given below, O is the centre of the circumcircle of triangle XYZ. Tangents at X and Y intersect at T. Given ∠XTY = 80° and ∠XOZ = 140°, calculate the value of ∠ZXY. (1994)
(b) In the figure (ii) given below, O is the centre of the circle and PT is the tangent to the circle at P. Given ∠QPT = 30°, calculate (i) ∠PRQ (ii) ∠POQ.


Answer 28


(a) Join OY, OX and OY are the radii of the circle
and XT and YT are the tangents to the circle.

Question 29

Two chords AB, CD of a circle intersect internally at a point P. If
(i) AP = cm, PB = 4 cm and PD = 3 cm, find PC.
(ii) AB = 12 cm, AP = 2 cm, PC = 5 cm, find PD.
(iii) AP = 5 cm, PB = 6 cm and CD = 13 cm, find CP.

Answer 29

In a circle, two chords AB and CD intersect
each other at P internally.


Question 30

(a) In the figure (i) given below, PT is a tangent to the circle. Find TP if AT = 16 cm and AB = 12 cm.

(b) In the figure given below, diameter AB and chord CD of a circle meet at P. PT is a tangent to the circle at T. CD = 7.8 cm, PD = 5 cm, PB = 4 cm. Find (i) AB. (ii)the length of tangent PT.


Answer 30


(a) PT is the tangent to the circle and AT is a secant.
PT² = TA × TB
Now TA = 16 cm, AB = 12 cm
TB = AT – AB = 16 – 12 = 4 cm
∴ PT² = 16 + 4 = 64 = (8)²
⇒ PT = 8 cm or TP = 8 cm
(b) PT is tangent and PDC is secant out to the circle
∴ PT² = PC × PD

Question 31

PAB is secant and PT is tangent to a circle
(i) PT = 8 cm and PA = 5 cm, find the length of AB.
(ii) PA = 4.5 cm and AB = 13.5 cm, find the length of PT.

Answer 31


∵ PT is the tangent and PAB is the secant of the circle.

Question 32

In the adjoining figure, CBA is a secant and CD is tangent to the circle. If AB = 7 cm and BC = 9 cm, then
(i) Prove that ∆ACD

∆DCB.
(ii) Find the length of CD.


Answer 32


In ∆ACD and ∆DCB
∠C = ∠C (common)
∠CAD = ∠CDB

Question 33

(a) In the figure (i) given below, PAB is secant and PT is tangent to a circle. If PA : AB = 1:3 and PT = 6 cm, find the length of PB.
(b) In the figure (ii) given below, ABC is an isosceles triangle in which AB = AC and Q is mid-point of AC. If APB is a secant, and AC is tangent to the circle at Q, prove that AB = 4 AP.


Answer 33

(a) In the figure (i),
PAB is secant and PT is the tangent to the circle.
PT² = PA × PB

Question 34

Two chords AB, CD of a circle intersect externally at a point P. If PA = PC, Prove that AB = CD.

Answer 34

Given: Two chords AB and CD intersect
each other at P outside the circle. PA = PC.

Question 35


(a) In the figure (i) given below, AT is tangent to a circle at A. If ∠BAT = 45° and ∠BAC = 65°, find ∠ABC.

(b) In the figure (ii) given below, A, B and C are three points on a circle. The tangent at C meets BA produced at T. Given that ∠ATC = 36° and ∠ACT = 48°, calculate the angle subtended by AB at the centre of the circle. (2001)


Answer 35


(a) AT is the tangent to the circle at A
and AB is the chord of the circle.

Question 36

In the adjoining figure ∆ABC is isosceles with AB = AC. Prove that the tangent at A to the circumcircle of ∆ABC is parallel to BC.


Answer 36


Given: ∆ABC is an isosceles triangle with AB = AC.
AT is the tangent to the circumcircle at A.

Question 37


If the sides of a rectangle touch a circle, prove that the rectangle is a square.

Answer 37

Given: A circle touches the sides AB, BC, CD and DA
of a rectangle ABCD at P, Q, R and S respectively.

Question 38


(a) In the figure (i) given below, two circles intersect at A, B. From a point P on one of these circles, two line segments PAC and PBD are drawn, intersecting the other circle at C and D respectively. Prove that CD is parallel to the tangent at P.

(b) In the figure (ii) given below, two circles with centers C, C’ intersect at A, B and the point C lies on the circle with centre C’. PQ is a tangent to the circle with centre C’ at A. Prove that AC bisects ∠PAB.


Answer 38


Given: Two circles intersect each other at A and B.
From a point P on one circle, PAC and PBD are drawn.
From P, PT is a tangent drawn. CD is joined.

Question 39


(a) In the figure (i) given below, AB is a chord of the circle with centre O, BT is tangent to the circle. If ∠OAB = 32°, find the values of x and y.

(b) In the figure (ii) given below, O and O’ are centers of two circles touching each other externally at the point P. The common tangent at P meets a direct common tangent AB at M. Prove that:
(i) M bisects AB (ii) ∠APB = 90°.


Answer 39


AB is a chord of a circle with centre O.
BT is a tangent to the circle and ∠OAB = 32°.
∴ In ∆OAB,
OA = OB (radii of the same circle)

MCQ Chapter – 15 Circle ML Aggarwal Solutions

Question 1

In the adjoining figure, O is the centre of the circle. If ∠OAB = 40°, then ∠ACB is equal to
(a) 50°
(b) 40°
(c) 60°
(d) 70°


Answer

In the given figure, O is the centre of the circle.
In ∆OAB,
∠OAB = 40°
But ∠OBA = ∠OAB = 40°

Question 2

ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140°, then ∠BAC is equal to
(a) 80°
(b) 50°
(c) 40°
(d) 30°

Answer


ABCD is a cyclic quadrilateral,
AB is the diameter of the circle circumscribing it
∠ADC = 140°, ∠BAC = Join AC

Question 3


In the adjoining figure, O is the centre of the circle. If ∠BAO = 60°, then ∠ADC is equal to
(a) 30°
(b) 45°
(c) 60°
(d) 120°


Answer


In the given figure, O is the centre of the circle ∠BAO = 60°

Question 4


In the adjoining figure, O is the centre of a circle. If the length of chord PQ is equal to the radius of the circle, then ∠PRQ is
(a) 60°
(b) 45°
(c) 30°
(d) 15°

Answer


In the given figure, O is the centre of the circle
Chord PQ = radius of the circle
∆OPQ is an equilateral triangle
∴∠POQ = 60°
Arc PQ subtends ∠POQ at the centre and
∴∠PRQ at the remaining part of the circle
∴∠PRQ = ∠POQ = x 60° = 30° (c)

Question 5

In the adjoining figure, if O is the centre of the circle then the value of x is
(a) 18°
(b) 20°
(c) 24°
(d) 36°


Answer

In the given figure, O is the centre of the circle.
Join OA.

Question 6

From a point which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is
(a) 60 cm²
(b) 65 cm²
(c) 30 cm²
(d) 32.5 cm²

Answer


Let point P is 13 cm from O, the centre of the circle
Radius of the circle (OQ) = 5 cm
PQ and PR are tangents from P to the circle
Join OQ and OR

Question 7

In the adjoining figure, PQ and PR are tangents from P to a circle with centre O. If ∠POR = 55°, then ∠QPR is
(a) 35°
(b) 55°
(c) 70°
(d) 80°


Answer

In the given figure,
PQ and PR are the tangents to the circle from a point P outside it

Question 8

In the adjoining figure, PA and PB are tangents from point P to a circle with centre O. If the radius of the circle is 5 cm and PA ⊥ PB, then the length OP is equal to
(a) 5 cm
(b) 10 cm
(c) 7.5 cm
(d) 5√2 cm

Answer

In the given figure,
PA and PB are tangents to the circle with centre O.
Radius of the circle is 5 cm, PA ⊥ PB.

Question 9

At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A is
(a) 4 cm
(b) 5 cm
(c) 6 cm
(d) 8 cm

Answer

AB is the diameter of a circle with radius 5 cm
At A, XAY is a tangent to the circle
CD || XAY at a distance of 8 cm from A
Join OC

Question 10


If radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle which is tangent to the other is
(a) 3 cm
(b) 6 cm
(c) 9 cm
(d) 1 cm

Answer


Radii of two concentric circles are 4 cm and 5 cm
AB is a chord of the bigger circle
which is tangent to the smaller circle at C.
Join OA, OC

Question 11


In the adjoining figure, O is the centre of a circle and PQ is a chord. If the tangent PR at P makes an angle of 50° with PQ, then ∠POQ is
(a) 100°
(b) 80°
(c) 90°
(d) 75°


Answer 11


In the given figure, O is the centre of the circle.
PR is tangent and PQ is chord ∠RPQ = 50°
OP is radius and PR is tangent to the circle

Question 12

In the adjoining figure, PA and PB are tangents to a circle with centre O. If ∠APB = 50°, then ∠OAB is equal to
(a) 25°
(b) 30°
(c) 40°
(d) 50°


Answer


In the given figure,
PA and PB are tangents to the circle with centre O.
∠APB = 50°
But ∠AOB + ∠APB = 180°
∠AOB + 50° = 180°
⇒ ∠AOB = 180° – 50° = 130°
In ∆OAB,
OA = OB (radii of the same circle)
∠OAB = ∠OBA
But ∠OAB + ∠OBA = 180° – ∠AOB
= 180° – 130° = 50°
∠OAB = = 25° (a)

Question 13


In the adjoining figure, sides BC, CA and AB of ∆ABC touch a circle at point D, E and F respectively. If BD = 4 cm, DC = 3 cm and CA = 8 cm, then the length of side AB is
(a) 12 cm
(b) 11 cm
(c) 10 cm
(d) 9 cm

Answer

In the given figure,
sides BC, CA and AB of ∆ABC touch a circle at D, E and F respectively.
BD = 4 cm, DC = 3 cm and CA = 8 cm

Question 14

In the adjoining figure, sides BC, CA and AB of ∆ABC touch a circle at the points P, Q and R respectively. If PC = 5 cm, AR = 4 cm and RB = 6 cm, then the perimeter of ∆ABC is
(a) 60 cm
(b) 45 cm
(c) 30 cm
(d) 15 cm

Answer

In the given figure, sides BC, CA and AB of ∆ABC
touch a circle at P, Q and R respectively
PC = 5 cm, AR = 4 cm, RB = 6 cm

Question 15

PQ is a tangent to a circle at point P. Centre of circle is O. If ∆OPQ is an isosceles triangle, then ∠QOP is equal to
(a) 30°
(b) 60°
(c) 45°
(d) 90°

Answer

PQ is tangent to the circle at point P centre of the circle is O.

Question 16

In the adjoining figure, PA and PB are tangents at points A and B respectively to a circle with centre O. If C is a point on the circle and ∠APB = 40°, then ∠ACB is equal to
(a) 80°
(b) 70°
(c) 90°
(d) 140°


Answer

In the given figure,
PA and PB are tangents to the circle at A and B respectively
C is a point on the circle and ∠APB = 40°
But ∠APB + ∠AOB = 180°

Question 17

In the adjoining figure, two circles touch each other at A. BC and AP are common tangents to these circles. If BP = 3.8 cm, then the length of BC is equal to
(a) 7.6 cm
(b) 1.9 cm
(c) 11.4 cm
(d) 5.7 cm


Answer

In the given figure, two circles touch each other at A.
BC and AP are common tangents to these circles
BP = 3.8 cm

Question 18

In the adjoining figure, if sides PQ, QR, RS and SP of a quadrilateral PQRS touch a circle at points A, B, C and D respectively, then PD + BQ is equal to
(a) PQ
(b) QR
(c) PS
(d) SR


Answer

In the given figure,
sides PQ, QR, RS and SP of a quadrilateral PQRS
touch a circle at the points A, B, C and D respectively
PD and PA are the tangents to the circle
∴ PA = PD …(i)
Similarly, QA and QB are the tangents
∴ QA = QB …(ii)
Now PD + BQ = PA + QA = PQ (a)
[From (i) and (ii)]

Question 19

In the adjoining figure, PQR is a tangent at Q to a circle. If AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to
(a) 20°
(b) 40°
(b) 35°
(d) 45°


Answer

In the given figure, PQR is a tangent at Q to a circle.
Chord AB || PR and ∠BQR = 70°
BQ is chord and PQR is a tangent
∠BQR = ∠A

Question 20

Two chords AB and CD of a circle intersect externally at a point P. If PC = 15 cm, CD = 7 cm and AP = 12 cm, then AB is
(a) 2 cm
(b) 4 cm
(c) 6 cm
(d) none of these


Answer 29

In the given figure,
two chords AB and CD of a circle intersect externally at P.
PC = 15 cm, CD = 7 cm, AP = 12 cm
Join AC and BD

Chapter-Test ML Aggarwal Class 10 for ICSE Maths Chapter -15 Circles

Question 1

In the figure given below ‘O’ is the centre of the circle. If QR = OP and ∠ORP = 20°. Find the value of ‘x ’ giving reasons.

Answer 1

Here, in ∆OPQ
OP = OQ = r
Also, OP = QR [Given]
OP = OQ = QR = r

In ∆OQR, OQ = QR
∠QOR = ∠ORP = 20°
And ∠OQP = ∠QOR + ∠ORQ
= 20° + 20°
= 40°
Again, in ∆ OPQ
∠POQ = 180° – ∠OPQ – ∠OQP
= 180°- 40° – 40°
= 100°
Now, x° + ∠POQ + ∠QOR = 180° [a straight angle]
x° + 100° + 20° = 180°
x° = 180° – 120° = 60°
Hence, the value of x is 60.

Question 2

(a) In the figure (i) given below, triangle ABC is equilateral. Find ∠BDC and ∠BEC.
(b) In the figure (ii) given below, AB is a diameter of a circle with centre O. OD is perpendicular to AB and C is a point on the arc DB. Find ∠BAD and ∠ACD


Answer 2


(a) ∆ABC is an equilateral triangle.

Question 3

(b) In the figure given below, AB is the diameter of the semi-circle ABCDE with centre O. If AE = ED and ∠BCD = 140°, find ∠AED and ∠EBD. Also Prove that OE is parallel to BD.



Answer 3

(a) We know that angle between the radius and the tangent at the point of contact is right angle.


Question 4

(a) In the figure (i) given below, O is the centre of the circle. Prove that ∠AOC = 2 (∠ACB + ∠BAC).
(b) In the figure (ii) given below, O is the centre of the circle. Prove that x + y = z.


Answer 4


(a) Given: O is the centre of the circle.
To Prove : ∠AOC = 2 (∠ACB + ∠BAC).
Proof: In ∆ABC,
∠ACB + ∠BAC + ∠ABC = 180° (Angles of a triangle)

Question 5

(a) In the figure (i) given below, AB is the diameter of a circle. If DC is parallel to AB and ∠CAB = 25°, find :
(i)∠ADC (ii) ∠DAC.

(b) In the figure give below sides AB and DC of a cyclic quadrilateral are produce to meet at a point P and the sides AD and BC are produce to meet at a point Q. If ∠ADC = 75° and ∠BPC = 50° , find ∠BAD and ∠CQD

Answer 5


(a) AB is diameter and DC || AB,
∠CAB = 25°, join AD,BD

Question 6

(a) In the figure (i) given below, ABCD is a cyclic quadrilateral. If AB = CD, Prove that AD = BC.
(b) In the figure (ii) given below, ABC is an isosceles triangle with AB = AC. If ∠ABC = 50°, find ∠BDC and ∠BEC.

Answer 6


(a) Given : ABDC is a cyclic quadrilateral AB = CD.
To Prove: AD = BC.

Question 7

A point P is 13 cm from the centre of a circle. The length of the tangent drawn from P to the circle is 12 cm. Find the distance of P from the nearest point of the circle.

Answer 7

Join OT, OP = 13 cm and TP = 12 cm

Question 8

Two circles touch each other internally. Prove that the tangents drawn to the two circles from any point on the common tangent are equal in length.

Answer 8


Given: Two circles with centre O and O’
touch each other internally at P.

Question 9

From a point outside a circle, with centre O, tangents PA and PB are drawn. Prove that
(i) ∠AOP = ∠BOP.
(ii) OP is the perpendicular bisector of the chord AB.

Answer 9


Given: From a point P, outside the circle with centre O.
PA and PB are the tangents to the circle,
OA, OB and OP are joined.

Question 10

(a) The figure given below shows two circles with centers A, B and a transverse common tangent to these circles meet the straight line AB in C. Prove that:
AP : BQ = PC : CQ.

(b) In the figure (ii) given below, PQ is a tangent to the circle with centre O and AB is a diameter of the circle. If QA is parallel to PO, prove that PB is tangent to the circle.


Answer 10

(a) Given: Two circles with centers A and B
and a transverse common tangent to these circles meets AB at C.

Question 11

In the figure given below, two circles with centers A and B touch externally. PM is a tangent to the circle with centre A and QN is a tangent to the circle with centre B. If PM = 15 cm, QN = 12 cm, PA = 17 cm and QB = 13 cm, then find the distance between the centers A and B of the circles.


Answer 11

In the given figure, two chords with centre A and B touch externally.
PM is a tangent to the circle with centre A
and QN is tangent to the circle with centre B.
PM = 15 cm, QN = 12 cm, PA = 17 cm, QB = 13 cm.
We have to find AB.

Question 12

Two chords AB, CD of a circle intersect externally at a point P. If PB = 7 cm, AB = 9 cm and PD = 6 cm, find CD.

Answer 12

∵ AB and CD are two chords of a circle
which intersect each other at P, outside the circle.

Question 13

(a) In the figure (i) given below, chord AB and diameter CD of a circle with centre O meet at P. PT is tangent to the circle at T. If AP = 16 cm, AB = 12 cm and DP = 2 cm, find the length of PT and the radius of the circle

(b) In the figure (ii) given below, chord AB and diameter CD of a circle meet at P. If AB = 8 cm, BP = 6 cm and PD = 4 cm, find the radius of the circle. Also, find the length of the tangent drawn from P to the circle. .


Answer 13

Given : (a) AB is a chord of a circle with centre O
and PT is tangent and CD is the diameter of the circle
which meet at P.
AP = 16 cm, AB = 12 cm, OP = 2 cm
∴PB = PA – AB = 16 – 12 = 4 cm
∵ABP is a secant and PT is tangent.
∴PT² = PA × PB.

Question 14

In the adjoining figure given below, chord AB and diameter PQ of a circle with centre O meet at X. If BX = 5 cm, OX = 10 cm and. the radius of the circle is 6 cm, compute the length of AB. Also, find the length of tangent drawn from X to the circle.


Answer 14


Chord AB and diameter PQ meet at X
on producing outside the circle

Question 15

(a) In the figure (i) given below, ∠CBP = 40°, ∠CPB = q° and ∠DAB = p°. Obtain an equation connecting p and q. If AC and BD meet at Q so that ∠AQD = 2 q° and the points C, P, B and Q are concyclic, find the values of p and q.
(b) In the figure (ii) given below, AC is a diameter of the circle with centre O. If CD || BE, ∠AOB = 130° and ∠ACE = 20°, find:
(i)∠BEC (ii) ∠ACB
(iii) ∠BCD (iv) ∠CED.


Answer 15


(a) (i) Given : ABCD is a cyclic quadrilateral.
Ext. ∠PBC = ∠ADC
⇒ 40° = ∠ADC

Question 17

(a) In the figure (i) given below, chords AB, BC and CD of a circle with centre O are equal. If ∠BCD = 120°, find
(i) ∠BDC (ii) ∠BEC
(iii) ∠AEC (iv) ∠AOB.
Hence Prove that AOAB is equilateral.


Answer 17


(a) In ∆BCD, BC = CD
∠CBD = ∠CDB
But ∠BCD + ∠CBD + ∠CDB = 180°
(∵ Angles of a triangle)

Question 18

(a) In the figure, (i) given below AB and XY are diameters of a circle with centre O. If ∠APX = 30°, find
(i) ∠AOX (ii) ∠APY (iii) ∠BPY (iv) ∠OAX.

(b) In the figure (ii) given below, AP and BP are tangents to the circle with centre O. If ∠CBP = 25° and ∠CAP = 40°, find :
(i) ∠ADB (ii) ∠AOB (iii) ∠ACB (iv) ∠APB.


Answer 18


(a) AB and XY are diameters of a circle with centre O.
∠APX = 30°.
To find :
(i) ∠AOX (ii) ∠APY
(iii) ∠BPY (iv) ∠OAX


The List of Important Formulas for Class 9 Circles is provided on this page. We have everything covered right from basic to advanced concepts in Circles. Make the most out of the Maths Formulas for Class 9 prepared by subject experts and take your preparation to the next level. Access the Formula Sheet of Circles Class 9 covering numerous concepts and use them to solve your Problems effortlessly.

Circle is the collection of all points in a plane, which are equidistant from a fixed point in the plane. The fixed point is called the centre O and the given distance is called the radius r of the circle.

Concentric circles: Circles having same centre and different radii are called concentric circles.

Arc: A continuous piece of a circle is called an arc of the circle.

Chord: A line segment joining any two points on a circle is called the chord of the circle.

  • Semicircle: A diameter of a circle divides it into two equal parts which are arc. Each of these two arcs is called semicircle.
  • Angle of semicircle is right angle.
  • If two arcs are equal, then their corresponding, chords are also equal.

Theorem 10.1: Equal chords of a circle subtend equal angle at the centre of the circle.
Theorem 10.2: If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal.
Theorem 10.3: The perpendicular drawn from centre to the chord of circle bisects the chord.
Theorem 10.4: The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord. Theorem 10.5: There is one and only one circle passing through three non-collinear points.
Theorem 10.6: Equal chords of circle are equidistant from centre.

Theorem 10.7: Chords equidistant from the centre of a circle are equal in length.

  • If two circles intersect in two points, then the line through the centres is perpendicular to the common chord.

Theorem 10.8: The angle subtended by an arc at the centre of circle is twice the angle subtended at remaining part of circumference.
Theorem 10.9: Any two angles in the same segment of the circle are equal.
Theorem 10.10: If a line segment joining two points subtends equal angles at two other points on the same side of the line containing the line segment, the four points lie on a circle (i.e., they are concyclic).

Cyclic Quadrilateral: If all the vertices of a quadrilateral lie on the circumference of circle, then quadrilateral is called cyclic.

Theorem 10.11: In a cyclic quadrilateral the sum of opposite angles is 180°.
Theorem 10.12: In a quadrilateral if the sum of opposite angles is 180°, then quadrilateral is cyclic.


How to calculate a perpendicular line

Let’s look at an example of how to use these equations. First, let’s assume you know the equation of the first line. It happens to be y=4x+5. Let’s also assume you know the x and y coordinates of a point that the perpendicular line passes through, say (4,5).

First, we need to calculate the slope. From the equation a = -1 / m we get a value of -1/4.

Next, we need to calculate the y-intercept of the new line using the equation b = y₀ + 1 * x₀ / m. From this, we get a value of 6.

Finally, we need to put this all together in the form: y=-1/4x + 6.


  1. The circle OJS is constructed so its radius is the difference between the radii of the two given circles. This means that JL = FP.
  2. We construct the tangent PJ from the point P to the circle OJS. This is done using the method described in Tangents through an external point.
  3. The desired tangent FL is parallel to PJ and offset from it by JL. Since PJLF is a rectangle, we need the best way to construct this rectangle. The method used here is to construct PF parallel to OL using the "angle copy" method as shown in Constructing a parallel through a point

As shown below, there are two such tangents, the other one is constructed the same way but on the bottom half of the circles.


Watch the video: GNTM 2. Κάθετη πασαρέλα σε ύψος 30 μέτρων. Η αντίδραση των κοριτσιών (November 2021).