NCERT Exemplar Class 10 Maths solutions Chapter 9 Circles 2021

Q: How many tangents can be drawn from a point inside the circle?

From any point inside the circle, we can never draw a tangent to the circle. Therefore, the answer to this question will be zero

Q: How many tangents can be drawn from a point outside the circle?

From any point outside the circle, we can draw two tangents to the circle. These two tangents will have equal lengths.

Q: Is the chapter Circles important for Board examinations?

Circles is an important chapter for Board examinations as it carries around 6-8% weightage of the whole paper.

Q: What is the question type distribution for the chapter of Circles in board examinations?

Generally, we get MCQ, Fill in the blanks, Short answer, and Long answer questions with a distribution of 1 question for each type. A thorough study and practice from NCERT exemplar Class 10 Maths solutions chapter 9 can help you score maximum marks for the same.

NCERT Exemplar Class 10 Maths solutions Chapter 9 Circles

Edited By Safeer PP | Updated on Sep 05, 2022 05:42 PM IST | #CBSE Class 10th

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NCERT exemplar Class 10 Maths solutions chapter 9 extends the learning of circles. In earlier classes, we have learned about circles; here, we will learn about tangents at any point of the circle. This chapter covers the definition of The NCERT exemplar Class 10 Maths chapter 9 solutions have been developed by our experienced Mathematics division subject matter experts to give students a quick and easy understanding of NCERT Class 10 Maths.

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Question:1

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 circle is
(A) 3 cm (B) 6 cm (C) 9 cm (D) 1 cm

Answer: (B) 6 cm

Solution
According to question

Here A is the center and AB = 4 cm radius of small circle and AD = 5 cm radius of large circle.
We have to find the length of CD
$\Delta$ ABD is a right angle triangle.
Hence use Pythagoras theorem in $\Delta$ ABD
$\left ( AD \right )^{2}= \left ( AB \right )^{2}+\left ( BD \right )^{2}$
$\left ( 5 \right )^{2}= \left ( 4 \right )^{2}+\left ( BD \right )^{2}$
$25-16= \left ( BD \right )^{2}$
$BD= \sqrt{9}= 3$
$BD= 3cm$
$CD= CB+BD$
$CD= CB+BD$
$CD= BD+BD= 2BD\; \; \left ( \because CB= BD \right )$
$CD= 2\times 3= 6cm$
$CD= 6cm$
Hence length of chord is 6 cm.

Question:2

In Figure, if ∠AOB = 125°, then ∠COD is equal to

(A) 62.5° (B) 45° (C) 35° (D) 55°

Answer: (D)
Solution

Given : ∠AOB = 125°
Let ∠COD = x°
As we know that the sum of opposite sides angles of a quadrilateral circumscribing a circle is equal to 180°.
Hence
∠AOB + ∠COD = 180°
125° + ∠COD = 180°
x° = 180° – 125° ( $\mathbb{Q}$ ∠COD = x°)
x° = 55°
Hence, ∠COD = 55°

Question:3

In Figure, AB is a chord of the circle and AOC is its diameter such that $<$ ACB = 50°. If AT is the tangent to the circle at the point A, then $\angle$ BAT is equal to

(A) 65° (B) 60° (C) 50° (D) 40°

Answer: (C)
Solution

Given $\angle$ ACB = 50°.
We know that the angle subtended by a diameter is right.
Hence $\angle$ B = 90°
We know that sum of interior angle of a triangle is 180°.
In $\bigtriangleup$ ABC
$\angle A+\angle B+\angle C= 180^{\circ}$
$\angle A+90^{\circ}+50^{\circ}= 180^{\circ}$
$\angle A=180^{\circ}-140^{\circ}$
$\angle A= 40^{\circ}$
That is $\angle CAB= 40^{\circ}$
$\angle CAT= \angle CAB+\angle BAT$
$90^{\circ}= 40^{\circ}+\angle BAT$ $\left (\because CA \perp AT \right )$
$\angle BAT= 90^{\circ}-40^{\circ}$
$\angle BAT= 50^{\circ}$

Question:4

From a point P 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:

(A) 60 cm²
Solution
According to question

Given OP = 13 cm
OQ = OR = radius = 5 cm
In $\bigtriangleup$ POQ, $\angle Q= 90^{\circ}$ ( $\because$ PQ is tangent)
Using Pythagoras theorem in $\bigtriangleup$ POQ
$\left ( PO \right )^{2}= \left ( PQ \right )^{2}+\left ( QO \right )^{2}$ ( because H² = B² + P²)
$\left ( 13 \right )^{2}= \left ( PQ \right )^{2}+\left ( 5 \right )^{2}$
$\left ( PQ\right )^{2}= 169-25$
$PQ= \sqrt{144}= 12$
$PQ= 12$
Area of $\bigtriangleup$ POQ = $\frac{1}{2}$ × perpendicular × base
$= \frac{1}{2}\times PQ\times OQ$
$= \frac{1}{2}\times 12\times 5= 30cm^{2}$
Area of quadrilateral = 2 × area of $\bigtriangleup$ POQ
= 2 × 30 = 60 cm²

Question:5

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:

(D) 8 cm
Solution
According to questions

Given AO = OB = 5 cm
Distance between XY and CD = 8 cm
Since D is the center of the circle and CD is chord. If we join OD it become the radius of circle that is OD = 5cm
AZ = 8 cm (given)
AZ = AO + OZ
8 = 5 + OZ
OZ = 3cm
In $\bigtriangleup$ ODZ, use Pythagoras theorem
$\left ( OD \right )^{2}= \left ( OZ \right )^{2}+\left ( ZD \right )^{2}$
$\left ( 5 \right )^{2}= \left ( 3 \right )^{2}+\left ( ZD \right )^{2}$
$\left ( ZD \right )^{2}= 25-9$
$ZD= \sqrt{16}= 4cm$
$CD= CZ+ZD$
$CD= ZD+ZD$ $\left ( \because CZ= ZD \right )$
$CD= 2ZD= 2\left ( 4 \right )= 8cm$
Length of chord CD = 8 cm

Question:6

In Figure, AT is a tangent to the circle with centre O such that OT = 4 cm and ∠OTA = 30°. Then AT is equal to

(A) 4 cm (B) 2 cm (C) $2\sqrt{3}cm$ (D) $4\sqrt{3}cm$

Answer: (C) $2\sqrt{3}cm$

Solution

Given ∠OTA = 30° , OT = 4cm
Join OA, since AT is tangent.
Hence it is perpendicular to OA
In $\bigtriangleup$ OAT
$\cos \theta = \frac{B}{H}$
$\cos 30^{\circ} = \frac{AT}{OT}$
$\frac{\sqrt{3}}{2}= \frac{AT}{4}\; \; \left ( \because \cos 30^{\circ}= \frac{\sqrt{3}}{2} \right )$
$AT= 4\times \frac{\sqrt{3}}{2}$
$AT= 2\sqrt{3}cm$

Question:7

In Figure, if O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50° with PQ, then ∠POQ is equal to

(A) 100° (B) 80° (C) 90° (D) 75°

Answer: (A) $100^{\circ}$

Solution
Given : $\angle QPR= 50^{\circ}$
We know that tangent is perpendicular to radius.
Hence $PR\perp PO$
$\angle$ OPR = $90^{\circ}$
$\angle$ OPR = $\angle$ OPQ + $\angle$ QPR
$90^{\circ}$ = $\angle$ OPQ + $50^{\circ}$
$\angle$ OPQ = $40^{\circ}$
$\angle$ OPQ = $\angle$ OQD $\left ( \because \, OP= OQ \, radius\, of\, circle\right )$
Hence , $\angle$ OPQ = $40^{\circ}$
We know that the sum of interior angles of a triangle is $180^{\circ}$
In $\bigtriangleup OPQ$
$\angle$ O+ $\angle$ Q + $\angle$ P = $180^{\circ}$
$\angle$ O +40+40 = 180
$\angle$ o = $100^{\circ}$
Hence $\angle$ OPQ = $100^{\circ}$

Question:8

In Figure, if PA and PB are tangents to the circle with centre O such that $\angle$ APB = 50°, then $\angle$ OAB is equal to

(A) 25° (B) 30° (C) 40° (D) 50°

Answer: (A) 25°

Solution
Given : $\angle$ APB = 50°
We know that length of tangents drawn from an external point is equal
Hence, PA = PB
Since, PA = PB
Let $\angle$ PAB = $\angle$ PBA = x⁰
In $\bigtriangleup$ PAB
$\angle$ p+ $\angle$ A+ $\angle$ B = $180^{\circ}$ ( $\because$ Sum of interior angles of a tangent is $180^{\circ}$ )
$50^{\circ}+x^{\circ}+x^{\circ}= 180^{\circ}$
$2x^{\circ}= 130^{\circ}$
$x^{\circ}= 65^{\circ}$
$\angle$ PAB = $\angle$ PBA = $65^{\circ}$
$\angle$ PAO = $90^{\circ}$ ( $\mathbb{Q}$ tangent is perpendicular to radius)
$\angle$ PAO = $\angle$ PAB + $\angle$ OAB
$90^{\circ}= 65^{\circ}+\angle OAB$
$\angle OAB= 90^{\circ}-65^{\circ}$
$\angle OAB= 25^{\circ}$

Question:9

If two tangents inclined at an angle 60° are drawn to a circle of radius 3 cm,then length of each tangent is equal to
(A) $\frac{3}{2}\sqrt{3}cm$ (B) 6 cm (C) 3 cm (D) $3\sqrt{3}cm$

Answer:

(D) $3\sqrt{3}cm$
Solution
According to question

Given OQ = OR = 3 cm (Radius)
$\angle P= 60^{\circ}$
Draw line OP which bisect $\angle P$ . That is $\angle OPQ= 30^{\circ}$
$\angle OPR= 30^{\circ}$
In $\bigtriangleup OPQ$
$\tan \theta = \frac{P}{B}$
$\tan 30^{\circ} = \frac{3}{PQ}$
$\frac{1}{\sqrt{3}}= \frac{3}{PQ}$
$PQ= 3\sqrt{3}$
Here PQ = PR
Hence , PQ = PR= $3\sqrt{3}$

Question:10

In Figure, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and $\angle$ BQR = 70°,then $\angle$ AQB is equal to

( A) 20° (B) 40° (C) 35° (D) 45°

Answer:

Answer(B) $40^{\circ}$
Solution
Given : $\angle BQR= 70^{\circ}$ , $AB\parallel PR$
Since DQ perpendicular to PR
$\angle DQR= 90^{\circ}$
$\angle DQR= \angle DQB+\angle BQR$
$90^{\circ}= \angle DQB+70^{\circ}$
$\angle DQB= 20^{\circ}$
Since DQ bisect $\angle AQB$
Hence , $\angle DQB=\angle DQA= 20^{\circ}$
$\angle AQB= \angle DQB+\angle DQA$
$= 20^{\circ}+20^{\circ}= 40^{\circ}$
$\angle AQB= 40^{\circ}$

Question:1

Write ‘True’ or ‘False’ and justify your answer in each of the following :
If a chord AB subtends an angle of $60^{\circ}$ at the centre of a circle, then angle between the tangents at A and B is also $60^{\circ}$ .

Answer:False

Solution

Here CA and CB are the two tangents which is drawn on chord AB and also we know that tangent and radius are perpendicular to each other.
i.e., $\angle CBO=\angle CAO= 90^{\circ}$
In quadrilateral, ABCD
$\angle A+\angle B+\angle C+\angle O= 360^{\circ}$
[ $\mathbb{Q}$ Sum of interior angles of a quadrilateral is $360^{\circ}$ ]
$90^{\circ}+90^{\circ}+\angle C+60^{\circ}= 360^{\circ}$
$\angle C= 360^{\circ}-60^{\circ}-90^{\circ}-90^{\circ}$
$\angle C= 120^{\circ}$
Here we conclude that angle between the tangents at A and B is $120^{\circ}$ .
Therefore the given statement is False.

Question:2

Write ‘True’ or ‘False’ and justify your answer in each of the following :The length of tangent from an external point on a circle is always greater than the radius of the circle.

Answer:

False
Solution
Case-I – When external point P is very close to circle

Here C is the center of circle. Let this radius is 5 and distance of Px and Py is 3.Or we can say that Case-I contradict the given statement.
Case-II – When external point P is far the circle.

Here radius of the given circle is 5 cm and let the length of Px and Py is 10 cm.
Hence according to Case-II and given statement is True.
Hence from the above two cases we conclude that the tangent’s length is depend on the distance of external point from the circle.
Therefore given statement is False because length of tangent from an external point of a circle may or may not be greater than the radius of the circle.
Therefore the given statement is False.

Question:3

Write ‘True’ or ‘False’ and justify your answer in each of the following :The length of tangent from an external point P on a circle with centre O is always less than OP.

Answer:

True
Solution

Here O is the center of given circle and PA is tangent which is drawn from an external point P. OA is radius of circle.We know that tangent and radius is always perpendicular to each other.
$\therefore \, \angle A= 90^{\circ}$
$\therefore$ $\bigtriangleup PAO$ is a right angle triangle Use Pythagoras theorem in $\bigtriangleup PAO$
$\left ( PO \right )^{2}= \left ( OA \right )^{2}+\left ( PA \right )^{2}$ …..(i)
From equation (i) we can say that PA is always less than PO.
In other words we can say that the length of hypotenuse is always greater than the length of perpendicular in a right angle triangle.
i.e., OP $>$ PA
Hence the given statement is True.

Question:4

Write ‘True’ or ‘False’ and justify your answer in each of the following :The angle between two tangents to a circle may be $0^{\circ}$ .

Answer:

True
Solution
It is possible that the angles of two line may be $0^{\circ}$ in only two conditions.
1. When lines are parallel
2.When both the lines are coincide.
Hence the given statement is True
This may be possible only when tangents are parallel or when both the tangents are coincide.

Question:5

Write ‘True’ or ‘False’ and justify your answer in each of the following :If angle between two tangents drawn from a point P to a circle of radius a and centre O is $90^{\circ}$ , then $OP= a\sqrt{2}$

Answer:

True
Solution

Given $\angle APB= 90^{\circ}$
Draw line OP from point O to P which bisect $\angle P$ .
i.e , $\angle OPB= 45^{\circ}$
In $\bigtriangleup OBP$
$\sin 45= \frac{OB}{OP}$
$\therefore \sin \theta = \frac{perpendicular}{hypotenuse}$

$\frac{1}{\sqrt{2}}= \frac{a}{OP}$
$OP= a\sqrt{2}$
Hence the given statement is True.

Question:6

Write ‘True’ or ‘False’ and justify your answer in each of the following :If angle between two tangents drawn from a point P to a circle of radius a and centre O is $60^{\circ}$ , then $OP= a\sqrt{3}$

Answer:

False
Solution

Given $\angle APB= 60^{\circ}$
Draw line OP from point O to P which bisect P. Which bisect $\angle P$
i.e, $\angle APO= 30^{\circ}$
In $\bigtriangleup OAP$
$\sin 30^{\circ}= \frac{OA}{OP}$
$\therefore \sin \theta = \frac{perpendicular}{hypotenuse}$

$\frac{1}{2}= \frac{a}{OP}$
OP = 2a
Hence the value of OP = 2a
Hence the given statement is False.

Question:7

Write ‘True’ or ‘False’ and justify your answer in each of the following :The tangent to the circumcircle of an isosceles triangle ABC at A, in which AB = AC, is parallel to BC.

Answer:

Answer: true
Solution

Given ABC is a isosceles triangle and AB = AC.
To Prove : $XY\parallel BC$
Proof : AB = AC (Given)
$\angle ABC= \angle ACB$
[ angle between chord of a circle and tangent is equal to angle made by chord in alternate segment]
Also $\angle XAB= \angle ABC$
$\Rightarrow \angle XAB= \angle BCA$
$\Rightarrow XAY\parallel BC$
i.e , $\Rightarrow XY\parallel BC$
Hence true

Question:8

Write ‘True’ or ‘False’ and justify your answer in each of the following
If a number of circles touch a given line segment PQ at a point A, then their centres lie on the perpendicular bisector of PQ.

Answer:

False
Solution
According to question.

Here C₁, C₂, C₃ are the circle with center O₁, O₂, O₃ respectively.
C₁, C₂, C₃ touches line PQ at point A. Here PQ is the tangent at each circle.
If we join O₁, O₂, O₃ to point A then the line is perpendicular to line PQ because if we draw a line from the center of circle at any point then the tangent at that point is perpendicular to the radius.
But here line joining the centers is not bisecting the line PQ because it depend on the length of PQ.
Hence the given statement is False.

Question:9

Write ‘True’ or ‘False’ and justify your answer in each of the following
If a number of circles pass through the end points P and Q of a line segment
PQ, then their centres lie on the perpendicular bisector of PQ.

Answer:

True
Solution
According to question.

Here C₁, C₂ circles pass through the point P and Q.
We know that the perpendicular bisector of chord of a circle is always passes through the center of the circle. Hence the perpendicular bisector of line PQ passes through the center of circles of C₁, C₂.
Hence the given statement is True.

Question:10

Write ‘True’ or ‘False’ and justify your answer in each of the following
AB is a diameter of a circle and AC is its chord such that $\angle$ BAC = 30°. If the tangent at C intersects AB extended at D, then BC = BD.

Answer:

True
Solution: First of all we solve the question according to give conditions. If we able to prove it then it will be true otherwise it will be false.
Given : $\angle$ BAC = $30^{\circ}$
Diagram : Construct figure according to given conditions then join BC and OC.

To Prove : BC = BD
Proof : $\angle$ BAC = $30^{\circ}$ (Given)
$\Rightarrow \, \angle BCD= 30^{\circ}$
[ $\because$ angle between chord and tangent id equal to the angle made by chord in alternate segment]
$\angle OCD= 90^{\circ}$
[ $\because$ Radius and tangent’s angle is always $90^{\circ}$ ]
In $\bigtriangleup$ OAC
OA = OC (both are radius of circle)
$\angle OCA= 30^{\circ}$
$\Rightarrow$ $\angle OCA= 30^{\circ}$ [opposite angles of an isosceles triangle is equal]
$\therefore \; \; \angle ACD= \angle ACO+\angle OCD$
$= 30+90= 120^{\circ}$
In $\bigtriangleup ACD$
$\angle CAD+\angle ADC+\angle DCA= 180^{\circ}$
[ $\because$ sum of interior angle of a trianglE $180^{\circ}$ ]
$30^{\circ}+\angle ADC+120^{\circ}= 180^{\circ}$
$\angle ADC= 180^{\circ}-120^{\circ}-30^{\circ}$
$< ADC=30^{\circ}$
In $\bigtriangleup$ BCD we conclude that
$\angle BCD= 30^{\circ}$ and $\angle ADC=30^{\circ}$
$\Rightarrow$ $BC= BD$ [ $\because$ sides which is opposite to equal angles is always equal]
Hence Proved.
Hence the given statement is true.

Question:1

Out of the two concentric circles, the radius of the outer circle is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle.

Answer:

Radius = 3 cm
Solution
According to question

Given : AC = 8 CM
OC = 5 CM
AC = AB+BC
8 = BC+BC ( $\because$ AB =BC)
BC = 4CM
In $\bigtriangleup OBC$ using Pythagoras theorem
$H^{2}+B^{2}+P^{2}$
$\left ( OC \right )^{2}= \left ( BC \right )^{2}+\left ( OB \right )^{2}$
$\left ( 5 \right )^{2}= \left ( 4 \right )^{2}+\left ( OB \right )^{2}$
$\left ( OB \right )^{2}= 25-16$
$OB= \sqrt{9}= 3cm$
Hence radius of inner circle is 3 cm.

Question:2

Two tangents PQ and PR are drawn from an external point to a circle with centre O. Prove that QORP is a cyclic quadrilateral.

Answer:

Solution
According to question

To Prove : QORP is a cyclic quadrilateral.
OQ $\perp$ PQ, OR $\perp$ PR ( $\because$ PQ, PR are tangents)
Hence, $\angle OQP+\angle ORP= 180^{\circ}\cdots (i)$
We know that sum of interior angles of quadrilateral is $360^{\circ}$
$\angle OPQ+\angle OPR+\angle ORP+\angle ROQ= 360^{\circ}$ [Given (i)]
$180^{\circ}+\angle QPR+\angle ROQ= 360^{\circ}$
$\angle QPR+\angle ROQ= 180^{\circ}$
Here we found that sum of opposite angles of quadrilateral is $180^{\circ}$
Hence QORP is a cyclic quadrilateral.
Hence proved

Question:3

If from an external point B of a circle with centre O, two tangents BC andBD are drawn such that $<$ DBC = $120^{\circ}$ , prove that BC + BD = BO, i.e., BO = 2BC.

Answer:

Solution
According to question

Given : $\angle DBC= 120^{\circ}$
To Prove : BC + BD = BO, i.e., BO = 2BC
Here OC $\perp$ BC, OD $\perp$ BD ( $\because$ BC, BD are tangents)
Join OB which bisect $\angle DBC$
In $\bigtriangleup ODB$
$\cos \theta = \frac{B}{H}$
$\cos 60^{\circ}= \frac{BD}{OB}$
$\frac{1}{2}= \frac{BD}{OB}$
$OB= 2BD$
$OB= 2BC$ $\left ( \because BD= BC \, \ \ Tangents \right )$
$OB= BC+BC$
$OB= BC+BD\; \; \left ( \because BC= BD \right )$
Hence Proved

Question:4

Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.

Answer:

Solution

Here PQ and PR are tangents and O is the center of circle.
Let us join OQ and OR.
Here $\angle OQP= \angle ORP= 90^{\circ}$
( $\because$ tangent from exterior point is perpendicular to the radius through the point of contact)
In $\bigtriangleup$ PQO and $\bigtriangleup$ PRO
$OQ= OR$ (Radius of circle)
$OP= OP$ (Common side)
Hence, $\bigtriangleup PQO\cong \bigtriangleup PRO$ [RHS interior]
Hence, $\angle RPO= \angle QPO$ [By CPCT]
Hence O lie on angle bisector of $\angle QPR$
Hence Proved.

Question:5

In Figure, AB and CD are common tangents to two circles of unequal radii. Prove that AB = CD.

Answer:

Solution
To Prove : AB = CD
Extend AB and CD then they meet at point E.

Here EA = EC ( $\because$ length of tangent drawn from same point is equal)
Also EB = ED ( $\because$ length of tangent drawn from some point is equal)
AB = AE - BE
CD = CE -DF
From equal axiom if equal are subtracted from equal then the result is equal.
$\therefore AB= CD$
Hence Proved

Question:6

In Figure, AB and CD are common tangents to two circles of unequal radii.

In above question, if radii of the two circles are equal, prove that AB = CD.

Answer:

Solution
To Prove AB = CD
According to question

It is given that radius of both circles are equal
Hence, OA = OC = PB = PD
Here, $\angle A= \angle B= \angle C= \angle D= 90^{\circ}$
( $\because$ tangent at any point is perpendicular to the radius at the point of contact)
Hence ABCD is a rectangle.
Opposite side of a rectangle are equal
$\therefore AB= CD$
Hence Proved.

Question:7

In Figure, common tangents AB and CD to two circles intersect at E. Prove that AB = CD.

Answer:

Solution
To Prove : AB = CD
We know that the length of tangents drawn from some point to a circle is equal.
So, EB = ED
AE = CE
Here AB = AE + EB
CD = CE + ED
From Euclid axiom is equals are added in equals then the result is also equal.
$\therefore$ AB = CD
Hence Proved

Question:8

A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ.

Answer:

Solution
According to question

To Prove : R bisects the arc PRQ
Here $\angle Q_{3}= \angle Q_{1}$ …..(i) ( $\because$ Alternate interior angles)
We know that angle between tangent and chord is equal to angle made by chord in alternate segment.
$\therefore \angle Q_{3}= \angle Q_{2}$ …..(ii)
From equation (i) and (ii)
$\angle Q_{1}= \angle Q_{2}$
We know that sides opposite to equal angles and equal.
$\therefore$ PR = QR
Hence R bisects the arc PRQ
Hence Proved

Question:9

Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.

Answer:

Solution
According to question

To Prove : $\angle Q_{1}= \angle Q_{2}$
X₁, Y₁, X₂, Y₂ are tangents at point A and B respectively.
Take a point C and join AC and AB.
Now $\angle Q_{1}= \angle C$ …..(i) (Angle in alternate segment)
Similarly,
$\angle Q_{2}= \angle C$ …..(ii)
From equation (i) and (ii)
$\angle Q_{1}= \angle Q_{2}= \angle C$ …..(iii)
From equation (3)
$\angle Q_{1}= \angle Q_{2}$
Hence Proved

Question:10

Prove that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A.

Answer:

Solution
According to question

Let us take a chord EF || XY
Here $\angle XAO= 90^{\circ}$
( $\because$ tangent at any point of the circle is perpendicular to the radius through the point of contact)
$\angle EGO= \angle XAO$ (Corresponding angles)
$\therefore \, \angle EGO= 90^{\circ}$
Thus AB bisects EF
Hence AB bisects all the chords which are parallel to the tangent at the point A.
Hence Proved

Question:1

If a hexagon ABCDEF circumscribe a circle, prove that AB + CD + EF = BC + DE + FA.

Answer:

Solution
Given ABCDEF hexagon circumscribe a circle.

To Prove : AB + CD + EF = BC + DE + FA
Proof : Here AR = AS [ $\because$ length of tangents drawn from a point are always equal]
Similarly,
BS = BT
CT = CU
DU = DV
EV = EW
FW = FR
Now AB + CD + EF = (AS + SB) + (CU + UD) + (EW + WF)
= (AR + BT) + (CT + DV) + (EV + FR)
= (AR + FR) + (BT + CT) + (DV + EV)
Now according to Euclid’s axiom when equals are added in equals then the result is also equal
$\Rightarrow$ AB + CD + EF = AF + BC + DE
i.e., AB + CD + EF = BC + DE + FA
Hence Proved

Question:2

Let s denote the semi-perimeter of a triangle ABC in which BC = a, CA = b, AB = c. If a circle touches the sides BC, CA, AB at D, E, F, respectively, prove that BD = s – b.

Answer:

Solution
Given : BC = a, CA = b, AB = c

Here, AF = AE = Z1 ( $\mathbb{Q}$ tangents drawn from an external point to the circle are equal in length)
$CE=CD=Z_{2}$
$BD=BF=Z_{3}$
Here AB + BC + CA = c + a + b
(AB + FB) + (BC + DC) + (CE + EA) = a + b + c
$\left ( AB+FB \right )+\left ( BC+DC \right )+\left ( CE+EA \right )= a+b+c$
$\left ( Z_{1} +Z_{3}\right )+\left ( Z_{3} +Z_{2}\right )+Z_{2} +Z_{1}= a+b+c$
$2\left ( Z_{1}+Z_{2} +Z_{3}\right )= a+b+c$
$2\left ( Z_{1}+Z_{2} +Z_{3}\right )= 2S$
$2\left ( Z_{1}+Z_{2} +Z_{3}\right )= 2S$
$\left ( a+b+c= 2s= perimeter\, of\, \bigtriangleup ABC \right )$
$Z_{1}+Z_{2}+Z_{3}= s$
$Z_{3}= s-\left ( Z_{1} +Z_{2}\right )$
- $\Rightarrow BD= s-b$ $\left [\because b= CE+EA= Z_{1}+Z_{2} \right ]$
Hence Proved

Question:3

From an external point P, two tangents, PA and PB are drawn to a circle with centre O. At one point E on the circle tangent is drawn which intersects PA and PB at C and D, respectively. If PA = 10 cm, find the perimeter of the triangle PCD.

Answer:

In this figure
CE = CA [ $\because$ Tangents from an external point to a circle are equal in length]
Similarly,
DE = DB and PB = PA
Perimeter of DPCD
= PC + CD + PD
= PC + CE + ED + PD ( $\because$ CD = CE + ED)
= PC + CA + DB + PD $\begin{bmatrix} \because CE= CA & \\ ED= DB& \end{bmatrix}$
= PA + PB [ $\because$ PC + CA = PA and DB + PD = PB]
= PA + PA [ $\because$ PB = PA]
= 2PA
= 2 × 10 [ $\because$ PA = 10]
= 20 cm

Question:4

If AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in Figure. Prove that $\angle$ BAT = $\angle$ ACB

Answer:

Solution
Here $\angle ABC= 90^{\circ}$ [ AC is a diameter line, $\therefore$ Angle in semi circle formed, is $90^{\circ}$ ]
In $\bigtriangleup$ ABC
$\angle CAB+\angle ABC+\angle BCA= 180^{\circ}$
[ $\because$ sum of interior angles of a triangle is $180^{\circ}$ ]
$\angle CAB+\angle BCA= 180^{\circ}-90^{\circ}$ ......(I)
We know that diameter of a circle is perpendicular to the tangent.
$\therefore CA\perp AT$
$\therefore \angle CAT= 90^{\circ}$
$\Rightarrow \angle CAB+\angle BAT= 90^{\circ}\cdots \left ( ii \right )$
Equate equation (i) and (ii) we get
$\angle CAB+\angle BAT= \angle CAB+\angle BCA$
$\angle BAT= \angle BCA$
Hence Proved

Question:5

Two circles with centres O and O' of radii 3 cm and 4 cm, respectively intersect at two points P and Q such that OP and O'P are tangents to the two circles. Find the length of the common chord PQ.

Answer:

4.8 cm
Solution

Given : Radii of two circles are OP = 3 cm and ${O}'$ and intersection point of two circles are P and Q. Here two tangents drawn at point P are OP and ${O}'$ P
$\therefore \: \angle P= 90^{\circ}$
$3^{2}-x^{2}= \left ( NP \right )^{2}$
Also apply Pythagoras theorem in $\bigtriangleup PN{O}'$ we get
$\left ( P{O}' \right )^{2}= \left ( PN \right )^{2}+\left (N {O}' \right )^{2}$
$\left (4 \right )^{2}= \left ( PN \right )^{2}+\left (5-x \right )^{2}$
$16= \left ( PN \right )^{2}+\left (5-x \right )^{2}$
$16- \left (5-x \right )^{2}= \left ( PN \right )^{2}\cdots \left ( ii \right )$
Equate equation (i) and (ii) we get
$9-x^{2}= 16-\left ( 5-x \right )^{2}$
$9-x^{2}-16+\left ( 25+x^{2}-10x \right )= 0$
$\left [ \because \left ( a-b \right )^{2}= a^{2}+b^{2}-2ab \right ]$
$-7-x^{2}+\left ( 25+x^{2}-10x \right )= 0$
$-7-x^{2}+25+x^{2}-10x= 0$
$18-10x= 0$
$18= 10x$
$\frac{18}{10}= x$
$x= 1\cdot 8$
Put x = 1.8 in equation (i) we get
$9-\left ( 1\cdot 8 \right )^{2}= NP^{2}$
$9-3\cdot 24= NP^{2}$
$5\cdot 76= NP^{2}$
$NP= \sqrt{5\cdot 76}$
$NP= 2\cdot 4$
$P\mathbb{Q}= 2\times PN= 2\cdot 24= 4\cdot 8 cm$

Question:6

In a right triangle ABC in which $\angle$ B = $90^{\circ}$ , a circle is drawn with AB as diameter intersecting the hypotenuse AC and P. Prove that the tangent to the circle at P bisects BC.

Answer:

Solution

Let O be the center of the given circle.
Suppose P meet BC at point R
Construction: Join point P and B
To Prove: BR = RC
Proof: $\angle ABC= 90^{\circ}$ [Given]
In $\bigtriangleup$ ABC
$\angle ABC+\angle BCA+\angle CAB= 180^{\circ}$ [ Sum of interior angle of a triangle is 180°]
$90^{\circ}+\angle 2+\angle 1= 180^{\circ}$
$\angle 1+\angle 2= 180^{\circ}-90^{\circ}$
$\angle 1+\angle 2= 90^{\circ}$
Also $\angle 4= \angle 1$ [ tangent and chord made equal angles in alternate segment]
$\Rightarrow \angle 4+\angle 2= 90^{\circ}\cdots \left ( i \right )$
$\angle APB= 90^{\circ}$ [ angle in semi circle formed is 90°]
$\angle APB+\angle BPC= 180^{\circ}$
$\angle BPC= 180^{\circ}-90^{\circ}$
$\angle BPC= 90^{\circ}$
$\angle 4+\angle 5= 90^{\circ}\cdots \left ( ii \right )$
Equal equation (i) and (ii) we get
$\angle 4+\angle 2= \angle 4+\angle 5$
$\angle 2= \angle 5$
$PR= RC \, \cdots \left ( iii \right )$ [Side opposite to equal angles are equal]
Also, PR = BR …..(iv) [ tangents drawn to a circle from external point are equal]
From equation (iii) and (iv)
BR = RC
Hence Proved

Question:7

In Figure, tangents PQ and PR are drawn to a circle such that $\angle$ RPQ = $30^{\circ}$ . A chord RS is drawn parallel to the tangent PQ. Find the $\angle$ RQS.

Answer:

Solution
In the given figure PQ and PR are two tangents drawn from an external point P.
$\therefore$ PQ = PR [ $\mathbb{Q}$ lengths of tangents drawn from on external point to a circle are equal]
$\Rightarrow \angle PQR=\angle QRP$ [ angels opposite to equal sides are equal]
In $\bigtriangleup$ PQR
$\angle PQR+\angle QRP+\angle RPQ= 180^{\circ}$
[ sum of angels of a triangle is 180°]
$\angle PQR+\angle PQR+\angle RPQ= 180^{\circ}$
$\left [ \angle PQR= \angle QRP \right ]$
$2\angle PQR+30^{\circ}= 180^{\circ}$
$\angle PQR= \frac{180^{\circ}-30^{\circ}}{2}$
$\angle PQR= \frac{150^{\circ}}{2}$
$\angle PQR= 75^{\circ}$
SR||OP (Given)
$\therefore$ $\angle SRQ= \angle RQP= 75^{\circ}$ [Alternate interior angles]
Also $\angle PQR= \angle QRS= 75^{\circ}$ [Alternate segment angles]
In $\bigtriangleup$ QRS
$\angle Q+\angle R+\angle S= 180^{\circ}$
$\angle Q+75^{\circ}+75^{\circ}= 180^{\circ}$
$\angle Q= 180^{\circ}-75^{\circ}-75^{\circ}$
$\angle Q= 30^{\circ}$
$\therefore \angle RQS= 30^{\circ}$

Question:8

AB is a diameter and AC is a chord of a circle with centre O such that $\angle$ BAC = $30^{\circ}$ . The tangent at C intersects extended AB at a point D. Prove that BC = BD.

Answer:

Solution

Given AB is a diameter and AC is a chord of circle with center O.
$\angle BAC= 30^{\circ}$
To Prove : BC = BD
Construction : Join B and C
Proof : $\angle BCD= \angle CAB$ [Angle is alternate segment]
$\angle BCD= 30^{\circ}\: \left [ Given \right ]$
$\angle BCD= 30^{\circ}\cdots \left ( i \right )$
$\angle ACB= 90^{\circ}$ [Angle in semi circle formed is 90°]
In $\bigtriangleup$ ABC
$\angle CAB+\angle ABC+\angle BCA= 180^{\circ}$ [Sum of interior angles of a triangle is 180°]
$30^{\circ}+\angle ABC+90^{\circ}= 180^{\circ}$
$\angle ABC= 180^{\circ}-90^{\circ}-30^{\circ}$
$\angle ABC= 60^{\circ}$
Also , $\angle ABC+\angle CBD= 180^{\circ}$ [Linear pair]
$\angle CBD= 180^{\circ}-60^{\circ}$
$\angle CBD= 120^{\circ}$
$\angle ABC= 60^{\circ}$
In $\bigtriangleup CBD$
$\angle CBD+\angle BDC+\angle DCB= 180^{\circ}$
$\angle BDC= 30^{\circ}\cdots \left ( ii \right )$
From equation (i) and (ii)
$\angle BCD= \angle BDC$
$\Rightarrow BC= BD$ [ $\because$ Sides opposite to equal angles are equal]
Hence Proved

Question:9

Prove that the tangent drawn at the mid-point of an arc of a circle is parallelto the chord joining the end points of the arc.

Answer:

Solution

Let mid-point of arc is C and DCE be the tangent to the circle.
Construction: Join AB, AC and BC.
Proof: In $\bigtriangleup$ ABC
AC = BC
$\Rightarrow \, \angle CAB= \angle CBA\: \cdots \left ( i \right )$
[ sides opposite to equal angles are equal]
Here DCF is a tangent line
$\therefore \; \; \angle ACD= \angle CBA$ [ angle in alternate segments are equal]
$\Rightarrow \; \angle ACD= \angle CAB$ …..(ii) [From equation (i)]
But Here $\angle ACD$ and $\angle CAB$ are alternate angels.
$\therefore$ equation (ii) holds only when AB||DCE.
Hence the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.
Hence Proved.

Question:10

In Figure, the common tangent, AB and CD to two circles with centres O and O' intersect at E. Prove that the points O, E, O' are collinear.

Answer:

Solution
Construction : Join AO and OS
${O}'D$ and ${O}'B$

In $\bigtriangleup E{O}'B$ and $\bigtriangleup E{O}'D$
${O}'D= {O}'B$ [Radius are equal]
${O}'E= {O}'E$ [Common side]
ED = EB [Tangent drawn from an external point to the line circle are equal to length]
$AE{O}'B\cong E{O}'D$ [By SSS congruence criterion]
$\Rightarrow \; \angle {O}'ED= \angle {O}'EB\, \cdots \left ( i \right )$
i.e., ${O}'E$ is bisector of $\angle BED$
Similarly OE is bisector of $\angle AEC$
In quadrilateral $DEB{O}'$
$\angle {O}'DE= \angle {O}'BE= 90^{\circ}$
$\Rightarrow \, \angle {O}'DE= \angle {O}'BE= 180^{\circ}$
$\therefore \angle DEB+\angle D {O}'B= 180^{\circ}\, \cdots \left ( ii \right )$ [ $\mathbb{Q}DEB{O}'$ is cyclic quadrilateral]
$\angle AED+\angle DEB= 180^{\circ}$ [ $\mathbb{Q}$ AB is a straight line]
$\angle AED+180^{\circ}-\angle D{O}'B= 180^{\circ}$ [From equation (ii)]
$\angle AED-\angle D{O}'B= 0$
$\angle AED-\angle D{O}'B\: \: \cdots \left ( iii \right )$
Similarly $\angle AED= \angle ADC\: \:\cdots \left ( iv \right )$
From equation (ii) $\angle DEB= 180^{\circ}-\angle D{O}'B$
Dividing both side by 2
$\frac{\angle DEB}{2}= \frac{180^{\circ}-\angle D{O}'B}{2}$
$\angle DE{O}'= 90^{\circ}-\frac{1}{2}\angle D{O}'B\: \: \cdots \left ( v \right )$
$\left [ \because \, \frac{\angle DEB}{2}= \angle DE{O}' \right ]$
Similarly $\angle AEC= 180^{\circ}-\angle AOC$
Dividing both side by 2
$\frac{1}{2}\angle AEC= 90^{\circ}-\frac{\angle AOC}{2}$
$\angle AEO= 90^{\circ}-\frac{1}{2}\angle AOC\; \cdots \left ( vi \right )$
$\left [ \because \frac{1}{2}\angle AEC= \angle AEO \right ]$
Now $\angle AEO+\angle AED+\angle DE{O}'$
$= 90-\frac{1}{2}\angle AOC+\angle AED+90^{\circ}-\frac{1}{2}\angle D{O}'B$
$= \angle AED+180^{\circ}-\frac{1}{2}\left ( \angle D{O}'B+\angle AOC \right )$
$= \angle AED+180^{\circ}-\frac{1}{2} \left [ \angle AED+\angle AED \right ]$ [from equation (iii) and (iv)]
$= \angle AED+180^{\circ}-\frac{1}{2} \left [ 2\angle AED \right ]$
$\angle AED+180^{\circ}-\angle AED$
$= 180^{\circ}$
$\therefore \: \angle AED+\angle DE{O}'+\angle AEO= 180^{\circ}$
So,OEO’ is straight line
$\therefore$ O, E and ${O}'$ are collinear.
Hence Proved

Question:11

In Figure. O is the centre of a circle of radius 5 cm, T is a point such that OT = 13 cm and OT intersects the circle at E. If AB is the tangent to the circle at E, find the length of AB.

Answer:

Answer $\frac{20}{3}\, cm$
Given: Radius = 5 cm, OT = 13 cm
$OP\perp PT$ [ PT is tangent]
Using Pythagoras theorem $\bigtriangleup$ OPT
$H^{2}+B^{2}+P^{2}$
$\left ( OT \right )^{2}= \left ( OP \right )^{2}+\left ( PT \right )^{2}$
$\left ( 13 \right )^{2}= \left ( 5 \right )^{2}+\left ( PT \right )^{2}$
$\left ( PT \right )^{2}= 169-25$
$PT= {\sqrt{144}}= 12\, cm$
PT and QT are tangents from same point
$\therefore$ PT = QT = 12 cm
AT = PT – PA
AT = 12 – PA ..…(i)
Similarly BT = 12 – QB ..…(ii)
Since PA, PF and BF, BQ are tangents from point A and B respectively.
Hence, PA = AE ..…(iii)
BQ = BE ..…(iv)
AB is tangent at point E
Hence OE $\perp$ AB
$\angle AET= 180^{\circ}-\angle AEO$ $\left ( \because \, \angle AEO= 90^{\circ} \right )$
$\angle AET= 180^{\circ}-90^{\circ}$
$\angle AET= 90^{\circ}$
ET = OT – OF
ET = 13 – 5
ET = 8 cm
In $\bigtriangleup$ AET, using Pythagoras theorem
$\left ( H \right )^{2}= \left ( B \right )^{2}+\left ( P \right )^{2}$
$\left ( AT \right )^{2}= \left ( AE \right )^{2}+\left ( ET \right )^{2}$
$\left ( 12-PA \right )^{2}= \left ( PA \right )^{2}+\left ( 8 \right )^{2}$ [using (i) and (ii)]
$144+\left ( PA \right )^{2}-24\left ( PA \right )-\left ( PA \right )^{2}-64= 0$
$\left [ \because \left ( a-b \right )^{2}= a^{2}+b^{2}-2ab \right ]$
$80-42\left ( PA \right )= 0$
$24\left ( PA \right )= 80$
$\left ( PA \right )= \frac{80}{24}= \frac{10}{3}\, cm$
$\therefore \; AE= \frac{10}{3}\, cm\, \left [ using\left ( iii \right ) \right ]$
Similiraly $BE= \frac{10}{3}\, cm$
$AB= AE+BE= \frac{10}{3}+\frac{10}{3}= \frac{20}{3}\, cm$

Question:12

The tangent at a point C of a circle and a diameter AB when extended intersect at P. If $\angle$ PCA = $110^{\circ}$ , find $\angle$ CBA [see Figure].

Answer:
Given : $\angle PCA= 110^{\circ}$
Here $\angle PCA= 90^{\circ}$ [ $\because$ PC is tangent]

$\angle PCA= \angle PCO+\angle OCA$
$110^{\circ}= 90^{\circ}+\angle OCA$
$\angle OCA= 20^{\circ}$
Here OC = OA (Radius)
$\therefore \: \angle OCA= \angle OAC= 20^{\circ}\cdots \left ( i \right )$ [ $\because$ Sides opposite to equal angles are equal]
We know that PC is tangent and angles in alternate segment are equal.
Hence $\angle BCP= \angle CAB= 20^{\circ}$
In $\bigtriangleup OAC$
$\angle O+\angle C+\angle A= 180^{\circ}$ [Interior angles sum of triangle is 180°]
$\angle O+20^{\circ}+20^{\circ}= 180^{\circ}$ [using (i)]
$\angle O= 180^{\circ}-40^{\circ}= 140^{\circ}$ …..(ii)
Here $\angle COB+\angle COA= 180^{\circ}$
$\angle COB= 180^{\circ}-140^{\circ}$ ( using (ii))
$\angle COB= 40^{\circ}$ …..(iii)
In $\bigtriangleup COB$
$\angle C+\angle O+\angle B= 180^{\circ}$ [using (iii)]
$90^{\circ}-20^{\circ}+40^{\circ}+\angle B= 180^{\circ}$
$\angle B= 180^{\circ}-110^{\circ}= 70^{\circ}$
Hence $\angle CBA= 70^{\circ}$

Question:13

If an isosceles triangle ABC, in which AB = AC = 6 cm, is inscribed in a circle of radius 9 cm, find the area of the triangle.

Answer:

Answer $8\sqrt{2}\, cm^{2}$
Solution
According to question

In $\bigtriangleup ABD$ and $\bigtriangleup ACO$
AB = AC [Given]
BO = CO [Radius]
AO = AO [Common side]
$\therefore \, \bigtriangleup ABO\cong \bigtriangleup ACO$ [By SSS congruence Criterion]
$\angle Q_{1}= \angle Q_{2}$ [CPCT]
In $\bigtriangleup ABD$ and $\bigtriangleup ACD$
AB = AC [given]
$\angle Q_{1}= \angle Q_{2}$
AD = AD [common side]
$\therefore \bigtriangleup ABD\cong \bigtriangleup ACD$ [By SAS congruence Criterion]
$\angle ADB= \angle ADC$ …..(i) [CPCT]
$\angle ADB= \angle ADC= 180^{\circ}$ …..(ii)
From (i) and (ii)
$\angle ADB= 90^{\circ}$
OA is a perpendicular which bisects chord BC
Let AD = x, then OD = 9 – x $\left ( \because OA= 9\, cm \right )$
Use Pythagoras in $\bigtriangleup$ ADC
$\left ( AC \right )^{2}= \left ( AD \right )^{2}+\left ( DC \right )^{2}$
$\left ( 6 \right )^{2}= x^{2}+\left ( DC \right )^{2}$
$\left ( AC \right )^{2}= 36-x^{2}$ …..(iii)
In $\bigtriangleup$ ODC using Pythagoras theorem
$\left ( OC \right )^{2}= \left ( OD \right )^{2}+\left ( DC \right )^{2}$
$\left ( DC \right )^{2}= 81-\left ( 9-x \right )^{2}$ …..(iv)
From (iii) and (iv)
$36-x^{2}= 81-\left ( 9-x \right )^{2}$
$36-x^{2}- 81+\left ( 81+x^{2} -18x\right )= 0$
$\left [ \because \left ( a-b \right )^{2} +a^{2}+b^{2}-2ab\right ]$
$36-x^{2}-81+81+x^{2}-18x= 0$
$18x= 36$
$x= 2$
i.e., AD = 2 cm, OD = 9 – 2 = 7 cm
Put value of x in (iii)
$\left ( DC \right )^{2}= 36-4$
$\left ( DC \right )^{2}= 32$
$DC= 4\sqrt{2}\, cm$
$BC= BD+DC$
$BC= 2DC$ $\left [ \because BD= DC \right ]$
$BC= 8\sqrt{2}\, cm$
Areao of $\bigtriangleup ABC= \frac{1}{2}\times base\times height$
$= \frac{1}{2}\times \times 8\sqrt{2}\times 2= 8\sqrt{2}\, cm$

Question:14

A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm.AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, find the perimeter of the $\bigtriangleup$ ABC.

Answer:

Answer 24 cm
Solution
Let us make figure according to question

Given : OA = 13 cm, Radius = 5 cm
Here AP = AQ (tangent from same point)
OP $\perp$ PA, OQ $\perp$ QA ( AP, AQ are tangents)
In $\bigtriangleup$ OPA using Pythagoras theorem
$\left ( H \right )^{2}= \left ( B \right )^{2}+\left ( P \right )^{2}$
$\left ( QP \right )^{2}= \left ( OP \right )^{2}+\left ( PA \right )^{2}$
$\left ( 13 \right )^{2}= \left ( 5 \right )^{2}+\left ( PA \right )^{2}$
$\left ( PA \right )^{2}= 169-25$
$PA= \sqrt{144}= 12\, cm$ ........(i)
Perimeter of $\bigtriangleup$ ABC = AB + BC + CA
= AB + BR + RC + CA
= AB + BD + CQ + CA
[ BP and BR are tangents from point B and CP and CQ are tangents from point C]
= AP + AQ [ AP = AB + BP, AQ = AC + CQ]
= AP + AP [ AP = AQ]
= 2AP
= 2 × 12 [using (i)]
= 24 cm

NCERT Exemplar Solutions Class 10 Maths Chapter 9 Important Topics:

Several theorems about a tangent to the circle and their proofs.
Find the number of tangents that can be drawn from any given point
In this chapter, students will learn that the tangent will be perpendicular to the line joining centre and the point of tangent on the circle.
Class 10 Maths NCERT exemplar chapter 9 solutions discusses the method that if we draw two tangents on diametrically opposite points of a circle we will observe that these two tangents are parallel.

NCERT Class 10 Exemplar Solutions for Other Subjects:

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NCERT Class 10 Maths Exemplar Solutions for Other Chapters:

Chapter 1 Real Numbers

Chapter 2 Polynomials

Chapter 3 Pair of Linear Equations in Two Variables

Chapter 4 Quadratic Equations

Chapter 5 Arithmetic Progressions

Chapter 6 Triangles

Chapter 7 Coordinate Geometry

Chapter 8 Introduction to Trigonometry & Its Equations

Chapter 10 Constructions

Chapter 11 Areas related to Circles

Chapter 12 Surface Areas and Volumes

Chapter 13 Statistics and Probability

Features of NCERT Exemplar Class 10 Maths Solutions Chapter 9:

These Class 10 Maths NCERT exemplar chapter 9 solutions provide an extension to the learning of circles done in Class 9. The chapter discusses tangents at any point of the circle along with the theorems such as the perpendicular tangent theorem. The chapter on Circles can be better understood and practiced using these NCERT exemplar Class 10 Maths chapter 9 solutions Circles and will be enough to solve other books such as A textbook of Mathematics by Monica Kapoor, NCERT Class 10 Maths, RD Sharma Class 10 Maths, RS Aggarwal Class 10 Maths et cetera.

NCERT exemplar Class 10 Maths solutions chapter 9 pdf download is a free and valuable feature. It provides the students with a pdf download facility to refer to the solutions in an offline environment while studying NCERT exemplar Class 10 Maths chapter 9.

Check Chapter-Wise Solutions of Book Questions

Chapter No.	Chapter Name
Chapter 1	Real Numbers
Chapter 2	Polynomials
Chapter 3	Pair of Linear Equations in Two Variables
Chapter 4	Quadratic Equations
Chapter 5	Arithmetic Progressions
Chapter 6	Triangles
Chapter 7	Coordinate Geometry
Chapter 8	Introduction to Trigonometry
Chapter 9	Some Applications of Trigonometry
Chapter 10	Circles
Chapter 11	Constructions
Chapter 12	Areas Related to Circles
Chapter 13	Surface Areas and Volumes
Chapter 14	Statistics
Chapter 15	Probability

Must, Read NCERT Solution Subject Wise

Check NCERT Notes Subject Wise

Also, Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

1. How many tangents can be drawn from a point inside the circle?

From any point inside the circle, we can never draw a tangent to the circle.

Therefore, the answer to this question will be zero

2. How many tangents can be drawn from a point outside the circle?

From any point outside the circle, we can draw two tangents to the circle.

These two tangents will have equal lengths.

3. Is the chapter Circles important for Board examinations?

Circles is an important chapter for Board examinations as it carries around 6-8% weightage of the whole paper.

4. What is the question type distribution for the chapter of Circles in board examinations?

Generally, we get MCQ, Fill in the blanks, Short answer, and Long answer questions with a distribution of 1 question for each type. A thorough study and practice from NCERT exemplar Class 10 Maths solutions chapter 9 can help you score maximum marks for the same.

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Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

sadhu ashram ramgart road aligarh

If you're looking for directions or steps to reach Sadhu Ashram on Ramgart Road in Aligarh, here’s how you can get there:

Steps to Reach Sadhu Ashram, Ramgart Road, Aligarh:

Starting Point:
- Determine your starting point in Aligarh or the nearby area.
Use Google Maps:
- Open Google Maps on your phone or computer.
- Enter "Sadhu Ashram, Ramgart Road, Aligarh" as your destination.
- Follow the navigation instructions provided by Google Maps.
By Local Transport:
- Auto-rickshaw/Taxi: Hire an auto-rickshaw or taxi and inform the driver about your destination. Most local drivers should be familiar with Sadhu Ashram.
- Bus: Check if there are any local buses that operate on Ramgart Road. Ask the bus conductor if the bus stops near Sadhu Ashram.
Landmarks to Look For:
- As you approach Ramgart Road, look for local landmarks that might help you confirm you’re on the right path, such as known shops, temples, or schools nearby.
Ask for Directions:
- If you're unsure, ask locals for directions to Sadhu Ashram on Ramgart Road. It's a known location in the area.
Final Destination:
- Once you reach Ramgart Road, Sadhu Ashram should be easy to spot. Look for any signage or ask nearby people to guide you to the exact location.

If you need detailed directions from a specific location or more information about Sadhu Ashram, feel free to ask

Read Complete Answer

sai raja 01 September,2024

show the previous year exams ?

Hello Aspirant, Hope your doing great, your question was incomplete and regarding what exam your asking.

Read Complete Answer

SWETCHCHA MISKA 10 July,2024

as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

Read Complete Answer

Manasvini Gupta 23 July,2024

i had cleared 10th from cbse in 2022 2 years ago now i wish to apply for 12th as private candidate in 2025. can i do so?

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

Read Complete Answer

Ashvadha 03 July,2024

i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

Read Complete Answer

sadafreen356 03 July,2024

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Central Board of Secondary Education Class 10th Examination

Applications Open

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Exemplar Class 10 Maths solutions Chapter 9 Circles

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NCERT Exemplar Solutions Class 10 Maths Chapter 9 Important Topics:

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