NCERT Solutions for Class 10 Maths Chapter 10 Circles

NCERT Solutions for Class 10 Maths Chapter 10 Circles

NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1

Ex 10.1 Class 10 Maths Question 1.
How many tangents can a circle have?
Solution:
There can be infinitely many tangents to a circle.

Ex 10.1 Class 10 Maths Question 2.
Fill in the blanks:
(i) A tangent to a circle intersects it in ………… point(s).
(ii) A line intersecting a circle in two points is called a ………… .
(iii) A circle can have ………………. parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called ……….. .
Solution:

NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1 PDF Q2

You can also download the free PDF of  Ex 10.1 Class 10 Circles NCERT Solutions or save the solution images and take the print out to keep it handy for your exam preparation.

 

Ex 10.1 Class 10 Maths Question 3.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is
(a) 12 cm
(b) 13 cm
(c) 8.5 cm
(d) √119 cm
Solution:
Ex 10.1 Class 10 Maths Solutions NCERT PDF Q3

Note: PQ = √119

Ex 10.1 Class 10 Maths Question 4.
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Solution:
Exercise 10.1 Class 10 Maths Solutions NCERT PDF Q4

Class 10 Maths Circles Mind Map

Introduction

A circle is a set of all points in a plane at a fixed distance from a fixed point in a plane. The fixed point is called the centre of the circle. The fixed distance is called the radius of the circle.

Line and a Circle

NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1 Mind Map 1
In Fig. (i), the line PQ and the circle have no common point. In this case, PQ is called a non-intersecting line with respect to the circle. In Fig. (ii), there are two common points A and B that the line PQ and the circle have. In this case, we call the line PQ a secant of the circle. In Fig. (iii), there is only one point A which is common to the line PQ and the circle. In this case, the line is called a tangent to the circle.

Tangent

A tangent to a circle is a straight line which touches the circle at only one point. The point where the tangent touches the circle is called point of contact of the tangent to the circle.
A tangent to a circle is a special case of a secant, when the two ends points of its corresponding chord coincides.
Theorem: Tangent at any point on a circle is perpendicular to the radius through the point of contact.
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1 Mind Map 2
CB is the tangent to the given circle touching at A and OA is the radius.
∴ ∠OAB = 90°
(i) At any point on the circle there can be one and only one tangent.
(ii) The line containing the radius through the point of contact is called the normal to the circle at the point.

Number of Tangents from a Point to Circle

(i) No tangent can be drawn from the point lying inside the circle, as shown in fig. (i)
(ii) One and only one tangent can be drawn from a point lying on the circle, as shown in fig. (ii)
(iii) Only two tangents can be drawn from an exterior point to a circle, as shown in fig. (iii)
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1 Mind Map 3

Length of a Tangent

The length of the segment of a tangent from an external point to the point of contact with the circle is called the length of the tangent

NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1 Mind Map 4
In the given figure, T1 and T2 are the points of contact of the tangents PT1 and PT2 respectively from the external point P.

Theorem Related to Length of Tangents From the External Points

The lengths of tangents drawn from an external point to a circle are equal.
i.e.,
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1 Mind Map 5
Here, PQ and PR are the two tangents drawn from P to the circle

NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2

Ex 10.2 Class 10 Maths Question 1.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(a) 7 cm
(b) 12 cm
(c) 15 cm
(d) 24.5 cm
Solution:
Ex 10.2 Class 10 Maths Solutions NCERT PDF Q1

Ex 10.2 Class 10 Maths Question 2.
In figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to
(a) 60°
(b) 70°
(c) 80°
(d) 90°
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 Q1
Solution:
Ex 10.2 Class 10 Maths Solutions NCERT PDF Q2

Ex 10.2 Class 10 Maths Question 3.
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to
(a) 50°
(b) 60°
(c) 70°
(d) 80°
Solution:
Ex 10.2 Class 10 Maths Solutions NCERT PDF Q3

Ex 10.2 Class 10 Maths Question 4.
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Solution:
Exercise 10.2 Class 10 Maths Solutions NCERT PDF Q4

Ex 10.2 Class 10 Maths Question 5.
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Solution:
Exercise 10.2 Class 10 Maths Solutions NCERT PDF Q5

Ex 10.2 Class 10 Maths Question 6.
The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Solution:
Exercise 10.2 Class 10 Maths Solutions NCERT PDF Q6

Ex 10.2 Class 10 Maths Question 7.
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Solution:
Circles Class 10 NCERT Solutions Ex 10.2 PDF Q7

Ex 10.2 Class 10 Maths Question 8.
A quadrilateral ABCD is drawn to circumscribe a circle (see figure). Prove that AB + CD = AD + BC.
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 Q8
Solution:
Circles Class 10 NCERT Solutions Ex 10.2 PDF Q8

Ex 10.2 Class 10 Maths Question 9.
In figure, XY and X’Y’ are two parallel tangents to a circle , x with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 Q9
Solution:
Chapter 10 Maths Class 10 NCERT Solutions Ex 10.2 PDF Q9

Ex 10.2 Class 10 Maths Question 10.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
Solution:
Chapter 10 Maths Class 10 NCERT Solutions Ex 10.2 PDF Q10

Ex 10.2 Class 10 Maths Question 11.
Prove that the parallelogram circumscribing a circle is a rhombus.
Solution:
Ch 10 Maths Class 10 NCERT Solutions Ex 10.2 PDF Q11

Ex 10.2 Class 10 Maths Question 12.
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see figure). Find the sides AB and AC.
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 Q12
Solution:
Ch 10 Maths Class 10 NCERT Solutions Ex 10.2 PDF Q12

Ex 10.2 Class 10 Maths Question 13.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution:
Ex 10.2 Class 10 Maths Solutions NCERT PDF Q13

Circles Class 10 Extra Questions Maths Chapter 10

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Important Questions for Class 10 Maths Chapter 10 Circles

Circles Class 10 Important Questions Very Short Answer (1 Mark)

Question 1.
In the given figure, O is the centre of a circle, AB is a chord and AT is the tangent at A. If ∠AOB = 100°, then calculate ∠BAT. (2011D)
Important Questions for Class 10 Maths Chapter 10 Circles 1
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 2
∠1 = ∠2
∠1 + ∠2 + 100° = 180°
∠1 + ∠1 = 80°
⇒ 2∠1 = 80°
⇒ ∠1 = 40°
∠1 + ∠BAT = 90°
∠BAT = 90° – 40° = 50°

Question 2.
In the given figure, PA and PB are tangents to the circle with centre O. If ∠APB = 60°, then calculate ∠OAB, (2011D)
Important Questions for Class 10 Maths Chapter 10 Circles 3
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 4
∠1 = ∠2
∠1 + ∠2 + ∠APB = 180°
∠1 + ∠1 + 60° = 180°
2∠1 = 180° – 60° = 120°
∠1 = 120∘/2 = 60°
∠1 + ∠OAB = 90°
60° +∠OAB = 90°
∠OAB = 90° – 60° = 30°

Question 3.
In the given figure, O is the centre of a circle, PQ is a chord and PT is the tangent at P. If ∠POQ = 70°, then calculate ∠TPQuestion (2011OD)
Important Questions for Class 10 Maths Chapter 10 Circles 5
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 6
∠1 = ∠2
∠1 + ∠2 + 70° = 180°
∠1 + ∠1 = 180° – 70°
2∠1 = 110° ⇒ ∠1 = 55°
∠1 + ∠TPQ = 90°
55° + ∠TPQ = 90°
⇒ ∠TPQ = 90° – 55° = 35°

Question 4.
A chord of a circle of radius 10 cm subtends a right angle at its centre. Calculate the length of the chord (in cm). (2014OD)
Solution:
AB= OA2 + OB2 …[Pythagoras’ theorem
Important Questions for Class 10 Maths Chapter 10 Circles 7
AB2 = 102 + 102
AB2 = 2(10)2
AB = 102 cm

Question 5.
In the given figure, PQ R is a tangent at a point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°. Find ∠PCA. (2016OD)
Important Questions for Class 10 Maths Chapter 10 Circles 8
Solution:
∠ACB = 90° …[Angle in the semi-circle
In ∆ABC,
∠CAB + ∠ACB + ∠CBA = 180°
30 + 90° + ∠CBA = 180°
∠CBA = 180° – 30° – 90° = 60°
∠PCA = ∠CBA …[Angle in the alternate segment
∴ ∠PCA = 60°

Question 6.
In the given figure, AB and AC are tangents to the circle with centre o such that ∠BAC = 40°. Then calculate ∠BOC. (2011OD)
Important Questions for Class 10 Maths Chapter 10 Circles 9
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 10
AB and AC are tangents
∴ ∠ABO = ∠ACO = 90°
In ABOC,
∠ABO + ∠ACO + ∠BAC + ∠BOC = 360°
90° + 90° + 40° + ∠BOC = 360°
∠BOC = 360 – 220° = 140°

Question 7.
In the given figure, a circle touches the side DF of AEDF at H and touches ED and EF produced at K&M respectively. If EK = 9 cm, calculate the perimeter of AEDF (in cm). (2012D)
Important Questions for Class 10 Maths Chapter 10 Circles 11
Solution:
Perimeter of ∆EDF
= 2(EK) = 2(9) = 18 cm

Question 8.
In the given figure, AP, AQ and BC are tangents to the circle. If AB = 5 cm, AC = 6 cm and BC = 4 cm, then calculate the length of AP (in cm). (2012OD)
Important Questions for Class 10 Maths Chapter 10 Circles 12
Solution:
2AP = Perimeter of ∆
2AP = 5 + 6 + 4 = 15 cm
AP = 15/2 = 7.5 cm

Question 9.
In the given figure, PA and PB are two tangents drawn from an external point P to a circle with centre C and radius 4 cm. If PA ⊥ PB, then find the length of each tangent. (2013D)
Important Questions for Class 10 Maths Chapter 10 Circles 13
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 14
Construction: Join AC and BC.
Proof: ∠1 = ∠2 = 90° ….[Tangent is I to the radius (through the point of contact
∴ APBC is a square.
Length of each tangent
= AP = PB = 4 cm
= AC = radius = 4 cm

Question 10.
In the given figure, PQ and PR are two tangents to a circle with centre O. If ∠QPR = 46°, then calculate ∠QOR. (2014D)
Important Questions for Class 10 Maths Chapter 10 Circles 15
Solution:
∠OQP = 900
∠ORP = 90°
∠OQP + ∠QPR + ∠ORP + ∠QOR = 360° …[Angle sum property of a quad.
90° + 46° + 90° + ∠QOR = 360°
∠QOR = 360° – 90° – 46° – 90° = 134°

Question 11.
In the given figure, PA and PB are tangents to the circle with centre O such that ∠APB = 50°. Write the measure of ∠OAB. (2015D)
Important Questions for Class 10 Maths Chapter 10 Circles 16Important Questions for Class 10 Maths Chapter 10 Circles 16
Solution:
PA = PB …[∵ Tangents drawn from external point are equal
∠OAP = ∠OBP = 90°
∠OAB = ∠OBA … [Angles opposite equal sides
∠OAP + ∠AOB + ∠OBP + ∠APB = 360° … [Quadratic rule
Important Questions for Class 10 Maths Chapter 10 Circles 17
90° + ∠AOB + 90° + 50° = 360°
∠AOB = 360° – 230°
= 130°
∠AOB + ∠OAB + ∠OBA = 180° … [∆ rule
130° + 2∠OAB = 180° … [From (i)
2∠OAB = 50°
⇒ ∠OAB = 25°

Question 12.
From an external point P, tangents PA and PB are drawn to a circle with centre 0. If ∠PAB = 50°, then find ∠AOB. (2016D)
Solution:
PA = PB …[∵ Tangents drawn from external point are equal
Important Questions for Class 10 Maths Chapter 10 Circles 18
∠PBA = ∠PAB = 50° …[Angles equal to opposite sides
In ∆ABP, ∠PBA + ∠PAB + ∠APB = 180° …[Angle-sum-property of a ∆
50° + 50° + ∠APB = 180°
∠APB = 180° – 50° – 50° = 80°
In cyclic quadrilateral OAPB
∠AOB + ∠APB = 180° ……[Sum of opposite angles of a cyclic (quadrilateral is 180°
∠AOB + 80o = 180°
∠AOB = 180° – 80° = 100°

Question 13.
In the given figure, PQ is a chord of a circle with centre O and PT is a tangent. If ∠QPT = 60°, find ∠PRQ. (2015OD)
Important Questions for Class 10 Maths Chapter 10 Circles 19
Solution:
PQ is the chord of the circle and PT is tangent.
∴ ∠OPT = 90° …[Tangent is I to the radius through the point of contact
Now ∠QPT = 60° … [Given
∠OPQ = ∠OPT – ∠QPT
⇒ ∠OPQ = 90° – 60° = 30°
In ∆OPQ, OP = OQ
∠OQP = ∠OPQ = 30° … [In a ∆, equal sides have equal ∠s opp. them
Now, ∠OQP + ∠OPQ + ∠POQ = 180°
∴ ∠POQ = 120° …[∠POQ = 180o – (30° + 30°)
⇒ Reflex ∠POQ = 360° – 120° = 240° …[We know that the angle subtended by an arc at the centre of a circle is twice the angle subtended by it at any point on the remaining part of the circle
∴ Reflex ∠POQ = 2∠PRO
⇒ 240° = 2∠PRQ
⇒ ∠PRQ = 240∘/2 = 120°

Question 14.
In the given figure, the sides AB, BC and CA of a triangle ABC touch a circle at P, Q and R respectively. If PA = 4 cm, BP = 3 cm and AC = 11 cm, find the length of BC (in cm). (2012D)
Important Questions for Class 10 Maths Chapter 10 Circles 20
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 21
AP = AR = 4 cm
RC = 11 – 4 = 7 cm
RC = QC = 7 cm
BQ = BP = 3 cm
BC = BQ + QC
= 3 + 7 = 10 cm

Question 15.
In a right triangle ABC, right-angled at B, BC = 12 cm and AB = 5 cm. Calculate the radius of the circle inscribed in the triangle (in cm). (2014OD)
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 22
AC2 = AB2 + BC2 …[Pythagoras’ theorem
= (5)2 + (12)2
AC2 = 25 + 144
AC = 169 = 13 cm
Area of ∆ABC = Area of ∆AOB + ar. of ∆BOC + ar. of ∆AOC
Important Questions for Class 10 Maths Chapter 10 Circles 23
60 = r(AB + BC + AC)
60 = r(5 + 12 + 13)
60 = 30r ⇒ r = 2 cm

Question 16.
Find the perimeter (in cm) of a square circum scribing a circle of radius a cm. (2011OD)
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 24
Radius = R
AB = a + a = 2a
∴ Perimeter = 4(AB)
= 4(2a)
= 8a cm

Question 17.
In the given figure, a circle is inscribed in a quadrilateral ABCD touching its sides AB, BC, CD and AD at P, Q, R and S respectively. If the radius DA of the circle is 10 cm, BC = 38 cm, PB = 27 cm and AD ⊥ CD, then calculate the length of CD. (2013OD)
Important Questions for Class 10 Maths Chapter 10 Circles 25
Solution:
Const. Join OR
Proof. ∠1 = ∠2 = 90° … [Tangent is ⊥ to the radius through the point of contact
∠3 = 90° …[Given
Important Questions for Class 10 Maths Chapter 10 Circles 26
∴ ORDS is a square.
DR = OS = 10 cm …(i)
BP = BQ = 27 cm …[Tangents drawn from an external point
∴ CQ = 38 – 27 = 11 cm
RC = CO = 11 cm …[Tangents drawn from an external point
DC = DR + RC = 10 + 11 = 21 cm …[From (i) & (ii)

Circles Class 10 Important Questions Short Answer-I (2 Marks)

Question 18.
Prove that the tangents drawn at the ends of a diameter of a circle are parallel. (2012OD)
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 27
Proof: ∠1 = 90° …(i)
∠2 = 90° …(ii)
∠1 = ∠2 … [From (i) & (ii)
But these are alternate interior angles
∴PQ || RS

Question 19.
In the figure, AB is the diameter of a circle with centre O and AT is a tangent. If ∠AOQ = 58°, find ∠ATQ.  (2015D)
Important Questions for Class 10 Maths Chapter 10 Circles 28
Solution:
∠ABQ = 1/2 ∠AOQ = 58∘/2 = 29°
∠BAT = 90° ….[Tangent is ⊥ to the radius through the point of contact
∠ATQ = 180° – (∠ABQ + ∠BAT)
= 180 – (29 + 90) = 180° – 119° = 61°

Question 20.
Two concentric circles are of radii 7 cm and r cm respectively, where r > 7. A chord of the larger circle, of length 48 cm, touches the smaller circle. Find the value of r. (2011D)
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 29
Given: OC = 7 cm, AB = 48 cm
To find: r = ?
∠OCA = 90° ..[Tangent is ⊥ to the radius through the point of contact
∴ OC ⊥ AB
AC = 1/2 (AB) … [⊥ from the centre bisects the chord
⇒ AC = 1/2 (48) = 24 cm
In rt. ∆OCA, OA2 = OC2 + AC2 … [Pythagoras’ theorem
r2 = (7)2 + (24)2
= 49 + 576 = 625
∴ r=  625= 25 cm

Question 21.
In the figure, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB. (2012D)
Important Questions for Class 10 Maths Chapter 10 Circles 30
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 31
Const.: Join OC
Proof: AB is a tangent to smaller circle and OC is a radius.
∴ ∠OCB = 90° … above theorem
In the larger circle, AB is a chord and OC ⊥ AB.
∴ AC = CB … [⊥ from the centre bisects the chord

Question 22.
In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6 cm, BC = 9 cm and CD = 8 cm. Find the length of side AD. (2011OD)
Important Questions for Class 10 Maths Chapter 10 Circles 32
Solution:
AB + CD = AD + BC
6 + 8 = AD + 9
14 – 9 = AD ⇒ AD = 5 cm

Question 23.
Prove that the parallelogram circumscribing a circle is a rhombus. (2012D, 2013D)
Solution:
Given. ABCD is a ॥gm.
To prove. ABCD is a rhombus.
Proof. In ॥gm, opposite sides are equal
Important Questions for Class 10 Maths Chapter 10 Circles 33
AB = CD
and AD = BC ..(i)
AP = AS …[Tangents drawn from an external point are equal in length
PB = BQ
CR = CO
DR = DS
By adding these tangents,
(AP + PB) + (CR + DR) = AS + BQ + CQ + DS
AB + CD = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC
AB + AB = BC + BC … [From (i)
2AB = 2 BC
AB = BC …(ii)
From (i) and (ii), AB = BC = CD = DA
∴ ॥gm ABCD is a rhombus.

Question 24.
In figure, a quadrilateral ABCD is drawn to circum- DA scribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points P, Q, RA and S respectively. Prove that: AB + CD = BC + DA. (2013 OD, 2016 OD)
Important Questions for Class 10 Maths Chapter 10 Circles 34
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 35
AP = AS ……(i) (Tangents drawn from an external point are equal in length
BP = BO …(ii)
CR = CQ ….(iii)
DR = DS ..(iv)
By adding (i) to (iv)
(AP + BP) + (CR + DR) = AS + BQ + CQ + DS
AB + CD = (BQ + CQ) + (AS + DS)
∴ AB + CD = BC + AD (Hence proved)

Question 25.
In the given figure, an isosceles ∆ABC, with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC. (2012D)
Important Questions for Class 10 Maths Chapter 10 Circles 36
Solution:
Given: The incircle of ∆ABC touches the sides BC, CA and AB at D, E and F F respectively.
Important Questions for Class 10 Maths Chapter 10 Circles 37
AB = AC
To prove: BD = CD
Proof: Since the lengths of tangents drawn from an external point to a circle are equal
∴ AF = AE … (i)
BF = BD …(ii)
CD = CE …(iii)
Adding (i), (ii) and (iii), we get
AF + BF + CD = AE + BD + CE
⇒ AB + CD = AC + BD
But AB = AC … [Given
∴ CD = BD

Question 26.
In Figure, a right triangle ABC, circumscribes a circle of radius r. If AB and BC are of lengths 8 cm and 6 cm respectively, find the value of r. (2012OD)
Important Questions for Class 10 Maths Chapter 10 Circles 38
Solution:
Const.: Join AO, OB, CO
Proof: area of ∆ABC
Important Questions for Class 10 Maths Chapter 10 Circles 39
From (i) and (ii), we get 12r = 24
∴ r = 2 cm

Question 27.
In the given figure, a circle inscribed in ∆ABC touches its sides AB, BC and AC at points D, E & F K respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, then find the lengths of AD, BE and CF. (2013D)
Important Questions for Class 10 Maths Chapter 10 Circles 40
Solution:
Let AD = AF = x
BD = BE = y …[Two tangents drawn from and an external point are equal
CE = CF = z
Important Questions for Class 10 Maths Chapter 10 Circles 41
AB = 12 cm …[Given
∴ x + y = 12 cm …(i)
Similarly,
y + z = 8 cm …(ii)
and x + z = 10 cm …(iii)
By adding (i), (ii) & (iii)
2(x + y + z) = 30
x + y + z = 15 …[∵ x + y = 12
z = 15 – 12 = 3
Putting the value of z in (ii) & (iii),
y + 3 = 8
y = 8 – 3 = 5
x + 3 = 10
x = 10 – 3 = 7
∴ AD = 7 cm, BE = 5 cm, CF = 3 cm

Question 28.
The incircle of an isosceles triangle ABC, in which AB = AC, touches the sides BC, CA and AB at D, E and F respectively. Prove that BD = DC. (2014OD)
Solution:
Given: The incircle of ∆ABC touches the sides BC, CA and AB at D, E and F respectively.
Important Questions for Class 10 Maths Chapter 10 Circles 42
AB = AC
To prove: BD = CD
Proof: AF = AE ..(i)
BF = BD …(ii)
CD = CE …(iii)
Adding (i), (ii) and (iii), we get
AF + BF + CD = AE + BD + CE
⇒ AB + CD = AC + BD
But AB = AC …[Given
∴ CD = BD

Question 29.
In the figure, a ∆ABC is drawn to circumscribe a circle of radius 3 cm, such that the segments BD and DC are respectively 6 cm 9 cm of lengths 6 cm and 9 cm. If the area of ∆ABC is 54 cm2, then find the lengths of sides AB and AC. (2011D, 2011OD, 2015 OD)
Important Questions for Class 10 Maths Chapter 10 Circles 43
Solution:
Given: OD = 3 cm; OE = 3 cm; OF = 3 cm ar(∆ABC) = 54 cm2
Important Questions for Class 10 Maths Chapter 10 Circles 44
Joint: OA, OF, OE, OB and OC
Let AF = AE = x
BD = BF = 6 cm
CD = CE = 9 cm
∴ AB = AF + BF = x + 6 …(i)
AC = AE + CE = x + 9 …(ii)
BC = DB + CD = 6 + 9 = 15 cm …(iii)
In ∆ABC,
Area of ∆ABC = 54 cm2 …[Given
ar(∆ABC) = ar(∆BOC) + ar(∆AOC) + ar(∆AOB)
Important Questions for Class 10 Maths Chapter 10 Circles 45

Question 30.
In the figure, a circle is inscribed in a ∆ABC, such that it touches the sides AB, BC and CA at points D, E and F respectively. If the lengths of sides AB, BC, and CA are 12 cm, 8 cm and 10 cm respectively, find the lengths of AD, BE and CF. (2016D)
Important Questions for Class 10 Maths Chapter 10 Circles 46
Solution:
AB = 12 cm, BC = 8 cm, CA = 10 cm
Important Questions for Class 10 Maths Chapter 10 Circles 47
As we know,
AF = AD
CF = CE
BD = BE
Let AD = AF = x cm
then, DB = AB – AD
= (12 – x) cm
∴ BE = (12 – x) cm ..[Tangents drawn from an external point are equal
Similarly,
CF = CE = AC – AF = (10 – x) cm
BC = 8 cm …[Given
⇒ BE + CE = 8 ⇒ 12 – x + 10 – x = 8
⇒ 22 – 8 = 2x ⇒ 2x = 14
∴ x = 7 ∴ AD = x = 7 cm
BE = 12 – x = 12 – 7 = 5 cm
CF = 10 – x = 10 – 7 = 3 cm

Question 31.
In Figure, common tangents AB and CD to the two circles with lo, centres O1 and O2 intersect at E. Prove that AB = CD. (2014OD)
Important Questions for Class 10 Maths Chapter 10 Circles 48
Solution:
EA = EC …(i) ….[Tangents drawn from an external point are equal
EB = ED …(ii)
EA + EB = EC + ED …[Adding (i) & (ii)
∴ AB = CD (Hence proved)

Question 32.
If from an external point P of a circle with centre O, two tangents PQ and PR are drawn such that ∠QPR = 120°, prove that 2PQ = PO. (2014D)
Solution:
∠OPQ = 1/2(∠QPR) ..[Tangents drawn from an external point are equal
1/2(120°) = 60° …[Tangent is ⊥ to the radius through the point of contact
∠OQP = 90°
In rt. ∆OQP, cos 60° = PQPO
1/2=PQ/PO ∴ 2PQ = PO

Question 33.
From a point T outside a circle of centre O, tangents TP and TQ are drawn to the circle. Prove that OT is the right bisector of line segment PQuestion (2015D)
Solution:
In ∆s’ TPC and TQC ….[Tangents drawn from an external point are equal
TP = TQ
TC = TC …[Common
∠1 = ∠2 …[TP and TQ are equally inclined to OT
∴ ∆TPC = ∆TQC … [SAS
∴ PC = QC …[CPCT
Important Questions for Class 10 Maths Chapter 10 Circles 49
∠3 = ∠4 …(i)
⇒ ∠3 + 24 = 180° … [Linear pair
⇒ ∠3 + ∠3 = 180°…[From (i)
⇒ 2∠3 = 180° ⇒ ∠3 = 90°
∴ ∠3 = ∠4 = 90°
∴ OT is the right bisector of PQuestion

Question 34.
In the figure, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If ∠PRQ = 120°, then prove that OR = PR + R. (2015OD)
Important Questions for Class 10 Maths Chapter 10 Circles 50
Solution:
Join OP and OO
∠OPR = 90°
PR = RQ … [Tangents drawn from an external point are equal
∠PRO = 1/2 ∠PRQ = 1/2 × 120° = 60°
Now, In ∆OPR,
Important Questions for Class 10 Maths Chapter 10 Circles 51
⇒ ∠OPR + ∠POR + ∠ORP = 180° …[∆ Rule
⇒ 90° + ∠POR + 60° = 180°
⇒ ∠POR + 150° = 180°
⇒ ∠POR = 30°
⇒ sin 30° = PR/OR ⇒ 1/2=PR/OR
⇒ OR = 2PR
⇒ OR = PR + QR (∵ PR = RQ) …(Hence proved)

Question 35.
In the figure, AP and BP are tangents to a circle with centre 0, such that AP = 5 cm and ∠APB = 60°. Find the length of chord AB. (2016D)
Important Questions for Class 10 Maths Chapter 10 Circles 52
Solution:
PA = PB …[Tangents drawn from an external point are equal
Given:
∠APB = 60°
∠PAB = ∠PBA … (i) …(Angles opposite to equal sides
In ∆PAB, ∠PAB + ∠PBA + ∠APB = 180° …[Angle-sum-property of a ∆
⇒ ∠PAB + ∠PAB + 60° = 180°
⇒ 2∠PAB = 180° – 60o = 120°
⇒ ∠PAB = 60°
⇒ ∠PAB = ∠PBA = ∠APB = 60°
∴ APAB is an equilateral triangle
Hence, AB = AP = 5 cm …[∵ All sides of an equilateral A are equal

Question 36.
In the figure, from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If OP = 2r, show that ∠OTS = ∠OST = 30°. (2016OD)
Important Questions for Class 10 Maths Chapter 10 Circles 53
Solution:
Let ∠TOP = θ …[Tangent is ⊥ to the radius through the point of contact
∠OTP = 90°
OT = OS = r … [Given
In rt. ∆OTP, cos θ = OT/OP
⇒ cos θ = r/2r ⇒ cos θ = 1/2
⇒ cos θ = cos 60° ⇒ θ = 60°
∴ ∠TOS = 60° + 60° = 120°
In ATOS,
∠OTS = ∠OST …[Angles opposite to equal sides
In ∠TOS,
∠TOS + ∠OTS + ∠OST = 180° … [Angle-sum-property of a ∆
120° + ∠OTS + ∠OTS = 180° … [From (i)
2∠OTS = 180° – 120°
∠OTS = 60°/2 = 30°
∴ ∠OTS = ∠OST = 30°

Question 37.
In the given figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Question If PA = 12 cm, QC = QD = 3 cm, then find PC + PD. (2017D)
Important Questions for Class 10 Maths Chapter 10 Circles 54
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 55
PA = PB = 12 cm …(i)
QC = AC = 3 cm …(ii)
QD = BD = 3 cm …(iii)
To find: PC + PD
= (PA – AC) + (PB – BD)
= (12 – 3) + (12 – 3) … [From (i), (ii) & (iii)
= 9 + 9 = 18 cm

Question 38.
In the figure, two circles touch each other at the point C. Prove that the common tangent to the circles at C, bisects the common tangent at P and Q. (2013 OD)
Important Questions for Class 10 Maths Chapter 10 Circles 56
Solution:
To prove: PR = RQ
Proof: PR = RC … (i)
QR = RC
From (i) and (ii), PR = QR (Hence proved)

Circles Class 10 Important Questions Short Answer-II (3 Marks)

Question 39.
In the figure, a circle is inscribed in a triangle PQR with PQ = 10 cm, QR = 8 cm and PR = 12 cm. Find the lengths of QM, RN and PI. (2012OD)
Important Questions for Class 10 Maths Chapter 10 Circles 57
Solution:
Let PL = PN = x cm
QL = QM = y cm
RN = MR = z cm
PQ = 10 cm = x + y = 10 …(i)
QR = 8 cm = y + z = 8 …(ii)
PR = 12 cm = x + z = 12 …(iii)
By adding (i), (ii) and (iii),
Important Questions for Class 10 Maths Chapter 10 Circles 58
We get,
⇒ 2x + 2y + 2z = 10 + 8 + 12
⇒ 2(x + y + z) = 30
⇒ x + y + z = 15
⇒ 10 + z = 15 … [From (i)
∴ z = 15 – 10 = 5 cm
From (ii)
y + 5 = 8
y = 8 – 5
y = 3 cm
From (iii)
x + 5 = 12
x = 12 – 5
x = 7 cm
∴ QM = 3 cm, RN = 5 cm, PL = 7 cm

Question 40.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. (2012D)
Solution:
1st method:
To prove. (i) ∠AOD + ∠BOC = 180°
(ii) ∠AOB + ∠COD = 180°
Proof. In ∆BPO and ∆BQO …[Tangents drawn from an external point are equal
Important Questions for Class 10 Maths Chapter 10 Circles 59
PO = 20 … [radii
BO = BO … [Common
∆BPO = ∆BQO … [SSS Congruency rule
∠8 = ∠1 …(i) (c.p.c.t.)
Similarly,
∠2 = ∠3, ∠4 = ∠5 and ∠6 = ∠7
∠1 + ∠2 + 23+ 24 + 25 + 26+ 27 + ∠8 = 360° …(Complete angles
∠1 + ∠2 + 22+ 25 + 25 + 26+ ∠6+ ∠1 = 360°
2(∠1 + ∠2 + 25 + 26) = 360°
∠BOC + ∠AOD = 180°…(i) [Proved part I
∠AOB + ∠BOC + ∠COD + ∠DOA = 360° …(Complete angles
∠AOB + ∠COD + 180o = 360° … [From (i)
∴ ∠AOB + ∠COD = 360° – 180o = 180° …(proved)

2nd method:
To prove:
(i) ∠6 + ∠8 = 180°
(ii) ∠5 + ∠7 = 180°
Proof. As AS and AP are tangents to the circle from a point A
∴ O lies on the bisector of ∠SAP
∴ ∠1 = 1/2 ∠BAD …(i)
Similarly BO, CO and DO are the bisectors of
∠ABC, ∠BCD and ∠ADC respectively. …(ii)
Important Questions for Class 10 Maths Chapter 10 Circles 60
∴ ∠1 + ∠4 + ∠3 + ∠2 =180°…(iii) ..[From (1) & (ii)
In ∆AOD, ∠1 + ∠2 + 26 = 180° …[Angle-sum-Prop. of a ∆
In ∆BOC, ∠3 + ∠4 + ∠8 = 180° …(v)
Adding (iv) and (v)
(∠1 + ∠2 + 23 + 24) + 26 + 28 = 180° + 180°
180° + 26 + 28 = 180° + 180° … [From (iii)
∴∠6 + 28 = 180°
Now ∠5 + ∠6 + ∠7 + ∠8 = 360° … (Complete angles
(∠5 + ∠7) + (∠6 + ∠8) = 360°
(∠5 + ∠7) + 180° = 360°
∠5 + ∠7 = 360° – 180° = 180°
∠5 + ∠7 = 180°

Question 41.
Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ. (2017D)
Solution:
We are given a circle with centre O, an external point T and two tangents TP and TQ to the circle, where P, Q are the points of contact (see Figure).
Important Questions for Class 10 Maths Chapter 10 Circles 61
We need to prove that:
∠PTQ = 2∠OPQ
Let ∠PTQ = θ
Now, TP = TQuestion ….[∵ Lengths of tangents drawn from an external pt. to a circle are equal
So, TPQ is an isosceles triangle.
Important Questions for Class 10 Maths Chapter 10 Circles 62

Circles Class 10 Important Questions Long Answer (4 Marks)

Question 42.
Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. (2011OD, 2012OD, 2013D, 2014OD, 2015D)
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 63
Given: XY is a tangent at point P to the circle with centre O.
To prove: OP ⊥ XY
Const.: Take a point Q on XY other than P and join to OQuestion
Proof: If point Q lies inside the circle, then XY will become a secant and not a tangent to the circle.
∴ OQ > OP
This happens with every point on the line XY except the point P.
OP is the shortest of all the distances of the point O to the points of XY
∴ OP ⊥ XY … [Shortest side is ⊥

Question 43.
Prove that the lengths of tangents drawn from an external point to a circle are equal. (2011D, 2012OD, 2013OD, 2014, 2015D & OD
2016D & OD, 2017D)
Solution:
Given: PT and PS are tangents from an external point P to the circle with centre O.
Important Questions for Class 10 Maths Chapter 10 Circles 64
To prove: PT = PS
Const.: Join O to P,
T & S
Proof: In ∆OTP and
∆OSP,
OT = OS …[radii of same circle
OP = OP …[circle
∠OTP – ∠OSP …[Each 90°
∴ AOTP = AOSP …[R.H.S
PT = PS …[c.p.c.t

Question 44.
Important Questions for Class 10 Maths Chapter 10 Circles 65
In the above figure, PQ is a chord of length 16 cm, of a circle of radius 10 cm. The tangents at P and Q intersect at a point T. Find the length of TP. (2014OD)
Solution:
TP = TQuestion .. [Tangents drawn from an external point
∆TPQ is an isosceles ∆ and TO is the bisector of ∠PTQ ,
OT ⊥ PQ …[Tangent is ⊥ to the radius through the point of contact
∴ OT bisects PQ
∴ PR = RQ = 16 = 8 cm …[Given
In rt. ∆PRO,
PR2 + RO2 = PO2 … [Pythagoras’ theorem
82 + RO2 = (10)2
RO2 = 100 – 64 = 36
∴ RO = 6 cm
Let TP = x cm and TR = y cm
Then OT = (y + 6) cm
In rt. ∆PRT, x2 = y2 + 82 …(i) …[Pythagoras’ theorem
In rt. ∆OPT,
OT2 = TP2 + PO2 …(Pythagoras’ theorem
(y + 6)2 = x2 + 102
y2 + 12y + 36 = y2 +64 + 100 …[From (i)
12y = 164 – 36 = 128 ⇒ y = 128/12=32/3
Putting the value of y in (i),
Important Questions for Class 10 Maths Chapter 10 Circles 66

Question 45.
In the figure, tangents PQ and PR are drawn from an external point P to a circle with centre O, such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQuestion Find ∠RQS. (2015D)
Important Questions for Class 10 Maths Chapter 10 Circles 67
Solution:
PR = PO …[∵ Tangents drawn from an external point are equal
⇒ ∠PRQ = ∠PQR …[∵ Angles opposite equal sides are equal
In ∆PQR,
Important Questions for Class 10 Maths Chapter 10 Circles 68
⇒ ∠PRQ + ∠RPQ + ∠POR = 180°…[∆ Rule
⇒ 30° + 2∠PQR = 180°
⇒ PQR=(18030)∘/2
= 75°
⇒ SR || QP and QR is a transversal
∵ ∠SRQ = ∠PQR … [Alternate interior angle
∴ ∠SRO = 75° …..[Tangent is I to the radius through the point of contact
⇒ ∠ORP = 90°
∴ ∠ORP = ∠ORQ + ∠QRP
90° = ∠ORQ + 75°
∠ORQ = 90° – 75o = 150
Similarly, ∠RQO = 15°
In ∆QOR,
∠QOR + ∠QRO + ∠OQR = 180° …[∆ Rule
∴∠QOR + 15° + 15° = 180°
∠QOR = 180° – 30° = 150°
⇒ ∠QSR = 1/2∠QOR
⇒ ∠QSR = 150∘/2 = 750 … [Used ∠SRQ = 75° as solved above
In ARSQ, ∠RSQ + ∠QRS + ∠RQS = 180° … [∆ Rule
∴ 75° + 75° + ∠RQS = 180°
∠RQS = 180° – 150o = 30°

Question 46.
Prove that 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. (2015OD)
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 69
B is the mid point of arc (ABC)
OA = OC …[Radius
OF = OF …[Common
∴ ∠1 = ∠2 …[Equal angles opposite equal sides
∴ ∆OAF = ∆OCF (SAS)
∴ ∠AFO = ∠CFO = 90° …[c.p.c.t
⇒ ∠AFO = ∠DBO = 90° …[Tangent is ⊥to the radius through the point of contact
But these are corresponding angles,
∴ AC || DE

Question 47.
In the figure, O is the centre of a circle of radius 5 cm. T is a point such that OT = 13 cm and OT intersects circle at E. If AB is a tangent to the circle at E, find the length of AB, where TP and TQ are two tangents to the circle. (2016D)
Important Questions for Class 10 Maths Chapter 10 Circles 70
Solution:
∠OPT = 90° …[Tangent is ⊥ to the radius through the point of contact
We have, OP = 5 cm, OT = 13 cm
In rt. ∆OPT,
OP2 + PT2 = OT? …[Pythagoras’ theorem
⇒ (5)2 + PT2 = (13)2
⇒ PT2 = 169 – 25 = 144 cm
⇒ PT = 144
= 12 cm
OP = OQ = OE = 5 cm … [Radius of the circle
ET = OT – OE
= 13 – 5 = 8 cm
Let, PA = x cm, then AT = (12 – x) cm
PA = AE = x cm …[Tangent drawn from an external point
In rt. ∆AET,
AE2 + ET2 = AT2 …(Pythagoras’ theorem
⇒ x2 + (8)2 = (12 – x)2
⇒ x2 + 64 = 144 + x2 – 24x
⇒ 24x = 144 – 64
x = 80/24=10/3 cm
AB = AE + EB = AE + AE = 2AE = 2x :
∴ AB = 2(10/3)=20/3cm=20/3cm
or 6.67 cm or 6.6 cm

Question 48.
In the figure, two equal circles, with centres 0 and O’, touch each other A at X. OO’ produced meets the circle with centre O’ at A. AC is tangent to the circle with centre O, at the point C. O’D is perpendicular to AC. Find the value of DO′/CO. (2016OD)
Important Questions for Class 10 Maths Chapter 10 Circles 71
Solution:
Given: Two equal circles, with centres O and O’, touch each other at point X. OO’ is produced to meet the circle with centre (at A. AC is tangent to the circle with centre O, at the point C. OʻD is perpendicular to AC.
To find: = DO′/CO
Proof: ∠ACO = 90° … [Tangent is ⊥ to the radius through the point of contact
In ∆AO’D and ∆AOC
∠O’AD = ∠OAC …(Common
∴ ∠ADO = ∠ACO …[Each 90°
∴ ∆AO’D ~ ∴AOC …(AA similarity
AO′/AO=DO′/CO … [In ~ As corresponding sides are proportional
r/3r=DO′/CO …[Let AO’ = O’X = OX = r ⇒ AO = r +r+ r = 3r
∴ DO′/CO=1/3

Question 49.
In the figure, l and m are two parallel tangents to a circle with centre O, touching the circle at A and B respectively. Another tangent at C intersects the line I at D and m at E. Prove that ∠DOE = 90°. (2013D)
Important Questions for Class 10 Maths Chapter 10 Circles 72
Solution:
Proof: Let I be XY and m be XY’
∠XDE + ∠X’ED = 180° … [Consecutive interior angles
Important Questions for Class 10 Maths Chapter 10 Circles 73
1/2XDE + 1/2∠X’ED =
1/2 (180°)
= ∠1 + ∠2 = 90° …[OD is equally inclined to the tangents
In ∆DOE, ∠1 + ∠2 + 23 = 180° …[Angle-sum-property of a ∆
90° + 23 = 180°
⇒ ∠3 = 180° – 90o = 90°
∴ ∠DOE = 90° …(proved)

Question 50.
In the figure, the sides AB, BC and CA of triangle ABC touch a circle with centre o and radius r at P, Q and R. respectively. (2013OD)
Prove that:
(i) AB + CQ = AC + BQ
(ii) Area (AABC) = 1/2 (Perimeter of ∆ABC ) × r
Important Questions for Class 10 Maths Chapter 10 Circles 74
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 75
Part I:
Proof: AP = AR …(i)
BP = BQ … (ii)
CQ = CR … (iii)
Adding (i), (ii) & (iii)
AP + BP + CQ
= AR + BQ + CR
AB + CQ = AC + BQ
Part II: Join OP, OR, OQ, OA, OB and OC
Proof: OQ ⊥ BC; OR ⊥ AC; OP ⊥ AB
ar(∆ABC) = ar(∆AOB) + ar(∆BOC) + ar (∆AOC)
Area of (∆ABC)
Important Questions for Class 10 Maths Chapter 10 Circles 76

Question 51.
In the figure, a triangle ABC is drawn to circumscribe a circle of radius 4 cm, such that the segments BD and DC are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC. (2014OD)
Important Questions for Class 10 Maths Chapter 10 Circles 77
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 78
Let AE = x ∴ AF = x
BC = 8 + 6 = 14 cm
AB = (x + 8) cm
AC = (x + 6) cm
∠1 = ∠2 = 23 = 90° …[ Tangent is ⊥ to the radius B [through the point of contact
Important Questions for Class 10 Maths Chapter 10 Circles 79
43x(x+14) = 2(2x + 28)
43x(x+14) = 2.2(x + 14)
3x(x + 14) = (x + 14)2 … [Squaring both sides
3x(x + 14) – (x + 14)2 = 0
(x + 14) [3x – (x + 14)] = 0
(x + 14) (2x – 14) = 0
x = -14 or x = 7
∴ x = 7 … [As side of ∆ cannot be -ve
∴ AB = x + 8 = 15 cm
and AC = x + 6 = 13 cm

Question 52.
Prove that the line segment joining the points of contact of two parallel tangents of a circle, passes through its centre. (2014 D)
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 80
Given: CD and EF are two C parallel tangents at points A and B of a circle with centre O.
To prove: AB passes through centre O or AOB is diameter of the circle.
Const.: Join OA and OB. Draw OM || CD.
Proof: ∠1 = 90° … (i)
…[∵ Tangent is I to the radius through the point of contact
OM || CD
∴ ∠1 + ∠2 = 180° …(Co-interior angles
90° + ∠2 = 180° …[From (i)
∠2 = 180° – 90o = 90°
Similarly, ∠3 = 90°
∠2 + ∠3 = 90° + 90° = 180°
∴ AOB is a straight line.
Hence AOB is a diameter of the circle with centre O.
∴ AB passes through centre 0.

CBSE Class 10 Maths Notes Chapter 10 Circles

Circle: A circle is a collection of all points in a plane which are at a constant distance from a fixed point.

Centre: The fixed point is called the centre.

Radius: The constant distance from the centre is called the radius.

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

Diameter: A chord passing through the centre of the circle is called diameter. It is the longest chord.

Tangent: When a line meets the circle at one point or two coincidings The line is known as points, a tangent.
The tangent to a circle is perpendicular to the radius through the point of contact.
⇒ OP ⊥ AB
Circles Class 10 Notes Maths Chapter 10 1
The lengths of the two tangents from an external point to a circle are equal.
⇒ AP = PB
Circles Class 10 Notes Maths Chapter 10 2

Length of Tangent Segment
PB and PA are normally called the lengths of tangents from outside point P.

Properties of Tangent to Circle

Theorem 1: Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Given: XY is a tangent at point P to the circle with centre O.
To prove: OP ⊥ XY
Construction: Take a point Q on XY other than P and join OQ
Proof: If point Q lies inside the circle, then XY will become a secant and not a tangent to the circle
OQ > OP
Circles Class 10 Notes Maths Chapter 10 3
This happens with every point on the line XY except the point P. OP is the shortest of all the distances of the point O to the points of XY
OP ⊥ XY …[Shortest side is the perpendicular]

Theorem 2: A line drawn through the end point of a radius and perpendicular to it, is the tangent to the circle.
Given: A circle C(O, r) and a line APB is perpendicular to OP, where OP is the radius.
To prove: AB is tangent at P.
Construction: Take a point Q on the line AB, different from P and join OQ.
Proof: Since OP ⊥ AB
OP < OQ ⇒ OQ > OP
Circles Class 10 Notes Maths Chapter 10 4
The point Q lies outside the circle.
Therefore, every point on AB, other than P, lies outside the circle.
This shows that AB meets the circle at point P.
Hence, AP is a tangent to the circle at P.

Theorem 3: Prove that the lengths of tangents drawn from an external point to a circle are equal
Given: PT and PS are tangents from an external point P to the circle with centre O.
To prove: PT = PS
Construction: Join O to P, T and S.
Circles Class 10 Notes Maths Chapter 10 5
Proof: In ∆OTP and ∆OSP.
OT = OS …[radii of the same circle]
OP = OP …[common]
∠OTP = ∠OSP …[each 90°]
∆OTP = ∆OSP …[R.H.S.]
PT = PS …[c.p.c.t.]

Note: If two tangents are drawn to a circle from an external point, then:

  • They subtend equal angles at the centre i.e., ∠1 = ∠2.
  • They are equally inclined to the segment joining the centre to that point i.e., ∠3 = ∠4.
    ∠OAP = ∠OAQ
    Circles Class 10 Notes Maths Chapter 10 6

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