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Class 10 Maths Chapter 10 Question Answers - Circles

SHORT ANSWER TYPE QUESTIONS

Q1. Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.

Sol.

Class 10 Maths Chapter 10 Question Answers - Circles

Let NM be a chord of a circle with centre C.
Let the tangents at M and N meet at O.
∵ OM is a tangent at M

∵ ∠OMC = 90° ...(1)
Similarly ∠ONC = 90° ...(2)
Since, CM = CN [Radii of the same circle]
∵ In D CMN, ∠1= ∠2
From (1) and (2), we have
∠OMC –∠1 = ∠ONC –∠2
⇒ ∠OML = ∠ONL

Thus, tangents make equal angles with the chord.

Q2. Two concentric circles have a common centre O. The chord AB to the bigger circle touches the smaller circle at P. If OP = 3 cm and AB = 8 cm then find the radius of the bigger circle.

Sol. ∵ AB touches the smaller circle at P.

Class 10 Maths Chapter 10 Question Answers - Circles

∴ OP ⊥ AB ⇒ ∠OPA = 90°
Now, AB is a chord of the bigger circle.
Since, the perpendicular from the centre to a chord, bisects the chord,
∴ P is the mid-point of AB

⇒ Class 10 Maths Chapter 10 Question Answers - Circles
In right ∆ APO, we have
AO= OP2 + AP2
⇒ AO2 = 32 + 42
⇒ AO2 = 9 + 16 = 25 = 52

⇒ Class 10 Maths Chapter 10 Question Answers - Circles

Thus, the radius of the bigger circle is 5 cm.

Q3. In the given figure, O is the centre of the circle and PQ is a tangent to it. If its circumference is 12π cm, then find the length of the tangent.

Sol. ∵ Circumference of the circle = 12π cm

Class 10 Maths Chapter 10 Question Answers - Circles

∴ 2π r =12π
[∵ r is the radius of the circle]

⇒  Class 10 Maths Chapter 10 Question Answers - Circles
⇒ Radius of the circle = 6 cm = OQ

Since a tangent to circle is perpendicular to the radius through the point of contact,
∴ ∠OQP = 90°
Now, in rt Δ OQP, we have:
OQ2 + QP2 = OP2
⇒ 62 + QP2 =102
⇒ QP2 = 102 − 62 = (10 − 6) (10 + 6) = 4 × 16 = 64 = 82

⇒  Class 10 Maths Chapter 10 Question Answers - Circles

Thus, the length of the tangent is 8 cm.

Q4. Given two concentric circles of radii 10 cm and 6 cm. Find the length of the chord of the larger circle which touches the other circle.

Sol. The chord AB touches the inner circle at P.

Class 10 Maths Chapter 10 Question Answers - Circles

∴ AB is tangent to the inner circle.
⇒ OP ⊥ AB

[∵ O is the centre and OP is radius through the point of contact P]
∴ ∠OPB = 90°.
Now, in right ∆ OPB, we have:
OP2 + PB2 = OB2
⇒ 62 + PB2 = 102
⇒ PB2 = 102 − 62
= (10 − 6) × (10 + 6)
⇒ PB2 = 4 × 16
⇒ PB2 = 64 = 82

⇒ Class 10 Maths Chapter 10 Question Answers - Circles

∵ The radius perpendicular to a chord bisects the chord.
∴ P is the mid-point of AB
∴ AB = 2 × PB = 2 × 8 = 16 cm.

Q5. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2 ∠OPQ.

Class 10 Maths Chapter 10 Question Answers - Circles

Sol. ∵ Tangent to a circle is perpendicular to the radius through the point of contact.
In quadrilateral. OPTQ,
∠OPT + ∠OQT + ∠POQ + ∠PTQ = 360°
or 90° + 90° + ∠POQ + ∠PTQ = 360°
⇒ ∠POQ + ∠PTQ = 360° − 90° − 90° = 180°   ...(1)
In Δ OPQ, ∠1 + ∠2 + ∠POQ = 180°   ...(2)

Since OP =OQ   [Radii of the same circle]
⇒ ∠1 = ∠2    [Angles opposite to equal sides]
∴  ∠OPT = 90° = ∠OQT
∴ From (2), we have
∠1 + ∠1 + ∠POQ = 180°
⇒ 2 ∠1 + ∠POQ = 180° ...(3)
From (1) and (3), we have
2 ∠1 + ∠POQ = ∠POQ + ∠PTQ
⇒ 2 ∠1= ∠PTQ
⇒ 2 ∠OPQ = ∠ PTQ.

Q6. In the figure, the incircle of ∆ ABC touches the sides BC, CA and AB at D, E and F respectively. If AB = AC, prove that BD = CD.

Class 10 Maths Chapter 10 Question Answers - Circles

Sol. Since the lengths of tangents drawn from an external point to a circle are equal,
∴ We have
AF = AE
BF = BD
CD = CE
Adding them, we get
(AF + BF) + CD = (AE + CE) + BD
⇒ AB + CD = AC + BD
But AB = AC  (Given)
∴ CD = BD.

Q7. A circle is touching the side BC of a D ABC at P and touching AB and AC produced at Q and R. Prove that:  Class 10 Maths Chapter 10 Question Answers - Circles

Sol. Since, the two tangents drawn to a circle from an external point are equal.
∵  AQ = AR ...(1)
Similarly, BQ = BP ...(2)
and CR = CP ...(3)

Class 10 Maths Chapter 10 Question Answers - Circles

Now, Perimeter of Δ ABC
= AB + BC + AC
= AB + (BP + PC) + AC
= AB + (BQ + CR) + AC
[From (2) and (3)]
= (AB + BQ) + (CR + AC)
= AQ + AR
= AQ + AQ [From (1)] = 2AQ
⇒ Class 10 Maths Chapter 10 Question Answers - Circles

Q8. In the given figure, the radii of two concentric circles are 13 cm and 8 cm. AB is diameter of the bigger circle. BD is the tangent to the smaller circle touching it at D. Find the length AD. 

Class 10 Maths Chapter 10 Question Answers - Circles

Sol. Join AE and OD
∵ ∠ODB = 90°
[BE is a tangent at D and OD is a radius] and ∠AEB = 90° [AB is diameter so ∠AEB is an angle in semicircle so ∠AEB = 90°]
∴ OD ║ ΑΕ and ∆BEA are similar by AA-Similarity

So Class 10 Maths Chapter 10 Question Answers - Circles
Class 10 Maths Chapter 10 Question Answers - Circles

Again in rt  Δ AED,

Class 10 Maths Chapter 10 Question Answers - Circles

Q9. In two concentric circles, a chord of the larger circle touches the smaller circle. If the length of this chord is 8 cm and the diameter of the smaller circle is 6 cm, then find the diameter of the larger circle.

Sol. Let the common centre be O. Let AB be the chord of the larger circle.

Class 10 Maths Chapter 10 Question Answers - Circles

∴ AB = 8 cm

And CD is the diameter of the smaller circle i.e., CD = 6 cm

⇒  Class 10 Maths Chapter 10 Question Answers - Circles
Join OA. D is the point of contact.
∴ OD ⊥ AB
⇒ D is the mid point of AB
⇒ AD = 4 cm Now, in right ΔADO, we have:
AO2 = AD2 + OD2
= 42 + 32 = 16 + 9 = 25 = 52
⇒ AO = 5 cm
⇒ 2AO = 2(5 cm) = 10 cm

∴ The diameter of the bigger circle is 10 cm.

Q10. In the following figure, PA and PB are two tangents drawn to a circle with centre O, from an external point P such that PA = 5 cm and ∠APB = 60°. Find the length of chord AB.

Class 10 Maths Chapter 10 Question Answers - Circles

Sol. Since the tangents to a circle from an external point are equal,
∴ PA = PB = 5 cm
In ΔPAB, we have
∠PAB = ∠PBA               [∵ PA = PB]
∴∠PAB + ∠PBA + ∠APB = 180°
⇒∠PAB + ∠PAB + 60° = 180°
⇒ 2 ∠PAB + 60° = 180°
⇒ 2 ∠PAB = 180°− 60°
= 120°
⇒ ∠PAB = 60°
⇒ Each angle of ∆PAB is 60°.
⇒ ΔPAB is an equilateral triangle. ∴
PA = PB
= AB = 5 cm
Thus, AB = 5 cm

Q11. In the following figure, AB is a chord of length 9.6 cm of a circle with centre O and radius 6 cm.

Class 10 Maths Chapter 10 Question Answers - Circles

The tangents at A and B intersect at P. Find the length PA.

Sol.

Class 10 Maths Chapter 10 Question Answers - Circles

Join OB.
Let PA = x cm and PR = y cm Since, OP is perpendicular bisector of AB

∴ AR = BR = 9.6/2 = 4.8 cm

Now, in rt ∆OAR, we have:
OA2 = OR2 + AR2
[By Pythagoras theorem]
⇒ OR2 = OA2 − AR2
= 62 − (4.8)2 = (6 − 4.8) × (6 + 4.8) = 1.2 × 10.8
⇒ = 12.96
OR = 3.6 cm.
Again, in right ΔOAP,
OP2 = AP2 + OA2
OP2 = (AR2 + PR2) + OA2
[∵ AP2 = AR2 + PR2]
⇒ (y + 3.6)2 = (4.8)2 + y2 + 62
⇒y2 + 12.96 + 7.2 y = 23.04 + y2 + 36
⇒ 7.2 y = 46.08

⇒  Class 10 Maths Chapter 10 Question Answers - Circles
⇒ PR = 6.4 cm
Now, AP2 = AP2 + PR2
= (4.8)2 + (6.4)2 = 23.04 + 40.96 = 64

⇒  Class 10 Maths Chapter 10 Question Answers - Circles


Q12. Two tangents PA and PB are drawn to a circle with centre O from an external point P. Prove  that ∠APB = 2∠OAB

Class 10 Maths Chapter 10 Question Answers - Circles

Sol. We have PA and PB, the tangents to the circle and O is the centre of the circle.
∴ PA = PB
⇒∠2 = ∠4  ...(1)

Class 10 Maths Chapter 10 Question Answers - Circles

Since the tangent is perpendicular to the radius through the point of contact,
∴∠OAP = 90°
⇒ ∠1 + ∠2 = 90° ...(2)
⇒ ∠2 = 90° − ∠1
Now, in ∆ABP, we have:
∴∠2 + ∠3 + ∠4 = 180°
⇒∠2 + ∠3 + ∠2 = 180° [From (1)]
⇒∠2 + ∠3 = 180°
⇒ 2 (90° − ∠1) + ∠3 = 180° [From (2)]
⇒ 180° − 2 ∠1 + ∠3 = 180°
⇒ 2 ∠1 = ∠3
⇒ ∠3 = 2∠1
⇒∠APB = 2∠OAB

Q13. ABC is an isosceles triangle, in which AB = AC, circumscribed about a circle. Show that BC is bisected at the point of contact.

Sol. We know that the tangents to a circle from an external point are equal.

Class 10 Maths Chapter 10 Question Answers - Circles

∴ AD = AF
Similarly,
BD = BE
and CE = CF
Since AB = AC [Given]
⇒ AB − AD = AC − AD
⇒ AB − AD = AC − AF [∵ AD = AF]
⇒ BD = CF ...(1)
But BF = BD and CF = CE
∴ From (1), we have:
BE = CE

Q14. If a, b, c are the sides of a right triangle where c is hypotenuse, prove that the radius r of the circle which touches the sides of the triangle is given by  Class 10 Maths Chapter 10 Question Answers - Circles

Sol. Here, a, b and c are the sides of rt D ABC such that BC = a, CA = b and AB = c Let the circle touches the sides BC, CA, AB at D, E and F respectively.

Class 10 Maths Chapter 10 Question Answers - Circles

= AE = AF and BD = BF
Also, CE = CD = r
∴ AF = b – r BF = a – r
Now, AB = c  ⇒  (AF + BF)
=(b – r) + (a – r)
⇒ c = b + a – 2r
⇒ 2r = a + b – c

Class 10 Maths Chapter 10 Question Answers - Circles

Q15. In a right Δ ABC, right angled at B, BC = 5 cm and AB = 12 cm. The circle is touching the sides of Δ ABC. Find the radius of the circle.

Sol. Let the circle with centre O and radius ‘r’ touches AB, BC and AC at P, Q, R, respectively.

Class 10 Maths Chapter 10 Question Answers - Circles

Now,
AR = AP
∵ AP = AB – BP = (12 – r) cm
∴ AR = (12– r)cm
Similarly, CR = (5 – r)cm
Now, using Pythagoras theorem in rt Δ ABC, we have
AC2 = AB2 + BC2
⇒ AC2 = 122 + 52
⇒ AC = 13 cm
But AC = AR + CR = (12 – r) + (5 – r)
⇒ (12 – r) + (5 – r) = 13 cm
⇒ 17 – 2r = 13 cm
⇒ 2 r = 17 – 13 = 4 cm
⇒ r = 4/2 = 2 cm

Thus, the radius of the circle is 2 cm.

Q16. Prove that the parallelogram circumscribing a circle is a rhombus.

Sol. Since ABCD is a ║ gm
∴ AB = CD
and AD = BC
∵ Tangents from an external point to a circle are equal,

Class 10 Maths Chapter 10 Question Answers - Circles

⇒(AP + PB) + (RC + DR)
  = (AS + DS) + (BQ + QC)
⇒ AB + CD = AD + BC
⇒ 2 AB =2 AD ⇒ AB = AD
⇒ AB = AD = CD = BC
i.e., ABCD is a rhombus.

Q17. In the following figure, OP is equal to diameter of the circle. Prove that ABP is an equilateral triangle.

Class 10 Maths Chapter 10 Question Answers - Circles

Sol. Since the tangent is perpendicular to the radius through the point of contact,
∴ ∠OAP = 90°

Class 10 Maths Chapter 10 Question Answers - Circles

Let us join AB and AC.
In right ∆OAP, OP is the hypotenuse and C is the mid point of OP.
[∵ OP is a diameter of the circle (given)]
∴ CA = CP = CO = Radius of the circle.
∴ ΔOAC is an equilateral triangle.
Since all angles in an equilateral triangle are 60°,
∴∠1 = 60°
Now, in ∆OAP, we have
∠1 + ∠OAP + ∠2 = 180°
⇒ 60° + 90° + ∠2 = 180°
⇒∠2 = 180° − 90° − 60° = 30°
Since PA and PB make equal angles with OP,
∴∠2= ∠3 ⇒∠3 = 30°
∴∠APB = ∠2 + ∠3
= 30° + 30° = 60°
Again, PA = PB.
⇒ In  ΔABP,∠4= ∠5
[Angles opposite to equal sides are equal]
Now, in ΔABP, ∠4 + ∠5 + ∠APB = 180°
⇒∠4 + ∠4 + ∠APB= 180°
⇒ 2∠4 + ∠60° = 180°
⇒ 2∠4 = 180° − 60° = 120°

Class 10 Maths Chapter 10 Question Answers - Circles

Since, ∠4 = 60° ∠5 = 60°
∵ ΔABP is an equilateral Δ.
∠APB = 60°

Q18. Prove that the angle between the two tangents to a circle drawn from an external point is supplementary to the angle subtended by the line segment joining the points of contact at the centre.

Or

Two tangents PA and PB are drawn from an external point P to a circle with centre O. Prove that AOBP is a cyclic quadrilateral.

Sol. 

Class 10 Maths Chapter 10 Question Answers - Circles

We have tangents PA and PB to the circle from the external point P. Since a tangent to a circle is perpendicular to the radius through the point of contact,
∴ ∠2 = 90° and ∠4 = 90°
Now, in quadrilateral OAPB,
∠1 + ∠2 + ∠3 + ∠4 = 360°
⇒∠1 + 90° + ∠3 + 90° = 360°
⇒∠1 + ∠3 = 360° − 90° − 90°
= 180°
i.e., ∠1 and ∠3 are supplementary angles.
⇒∠ AOB and ∠APB are supplementary
⇒ AOBP is a cyclic quadrilateral.

Q19. Two equal circles, with centres O and O′, touch each other 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  Class 10 Maths Chapter 10 Question Answers - Circles

Class 10 Maths Chapter 10 Question Answers - Circles

Sol. AC is tangent to circle with centre O at C (given)
∠ACO = 90°  ⇒ ∆ ACO is a rt Δ
∠ADO′ = 90°   [ ä O′D ⊥ AC] ⇒ Δ
ADO′ is a rt Δ

Class 10 Maths Chapter 10 Question Answers - Circles

Class 10 Maths Chapter 10 Question Answers - Circles

But, AO′ = r,     O′X = r  and  OX = r ⇒ AO = 3r

Class 10 Maths Chapter 10 Question Answers - Circles
From (1) and (2), we get  Class 10 Maths Chapter 10 Question Answers - Circles

Q20. Out of 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.

Class 10 Maths Chapter 10 Question Answers - Circles

Sol. Let the given chord AC of the larger circle touch the smaller circle at L. ∵ AC is a tangent at L to the smaller circle with centre O
∴ OL ⊥ AC
Also AC is a chord of the bigger circle

∴  Class 10 Maths Chapter 10 Question Answers - Circles

Now, in rt. ΔOAL,
OL2 = OA2 – AL2

or OL2 = 52 – 42
= (5 + 4) (5 – 4)
= 9 × 1 = 9

⇒  Class 10 Maths Chapter 10 Question Answers - Circles

Thus, the radius of the inner circle is 3 cm.

Q21. In the figure, O is the centre of a circle of radius 5cm. T is a point such that OT = 13cm 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. 

Class 10 Maths Chapter 10 Question Answers - Circles

Sol. O is centre of the circle and PT is a tangent to circle
∴ ∠ OPT = 90°  ⇒ ∆ OPT is a rt ∆ using Pythagoras theorem
OT2 = OP2 + PT2   or 132 = 52 + PT2
⇒ PT2 =132 – 52  
⇒ PT =  Class 10 Maths Chapter 10 Question Answers - Circles

Let AP = AE = x
[Tangent to a circle from an external point are equal]
⇒ AT = PT – AP = (12 – x) cm
[∵ AB is a tangent to the circle at E and OE is a radius]
∴ ∠ OEA = 90° ⇒ ∠AET  = 90°
Δ AET is a rt ∆
⇒ AT2 = AE2 + ET2
or (12 − x)2 = x2 + (13 − 5)2
⇒  144− 24x + x2 = x2 + 64 or  24x = 80

Class 10 Maths Chapter 10 Question Answers - Circles

The document Class 10 Maths Chapter 10 Question Answers - Circles is a part of the Class 10 Course Mathematics (Maths) Class 10.
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FAQs on Class 10 Maths Chapter 10 Question Answers - Circles

1. What is a circle?
Ans. A circle is a closed curve in which all points are equidistant from a fixed point called the center.
2. How is the radius of a circle related to its diameter?
Ans. The radius of a circle is half of its diameter. In other words, if the diameter of a circle is d units, then its radius is d/2 units.
3. What is the formula to find the circumference of a circle?
Ans. The formula to find the circumference of a circle is C = 2πr, where C is the circumference and r is the radius of the circle.
4. How can we find the area of a circle?
Ans. The area of a circle can be found using the formula A = πr^2, where A is the area and r is the radius of the circle.
5. How is the diameter of a circle related to its circumference?
Ans. The diameter of a circle is the distance across it passing through the center. It is directly related to the circumference of the circle. The circumference of a circle is equal to π times the diameter. So, if the diameter of a circle is d units, then its circumference is πd units.
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