KENDRIYA VIDYALAYA EMBASSY OF INDIA, KATHMANDU, NEPAL

PRACTICE PAPER 10 - CHAPTER 10 CIRCLES (2025-26)

(ANSWERS)

SUBJECT: MATHEMATICS MAX. MARKS : 40

CLASS : X DURATION : 1½ hrs
General Instructions:
(i). All questions are compulsory.
(ii). This question paper contains 20 questions divided into five Sections A, B, C, D and E.
(iii). Section A comprises of 10 MCQs of 1 mark each. Section B comprises of 4 questions of 2 marks
each. Section C comprises of 3 questions of 3 marks each. Section D comprises of 1 question of 5
marks each and Section E comprises of 2 Case Study Based Questions of 4 marks each.
(iv). There is no overall choice.
(v). Use of Calculators is not permitted

SECTION – A
Questions 1 to 10 carry 1 mark each.

1. In the given figure, AB and AC are tangents to the circle with centre O such that BAC = 40°,
then BOC is equal to

(a) 40° (b) 50° (c) 140° (d) 150°

Ans. (c) 140°
In quadrilateral ABOC
ABO + BOC + OCA + BAC = 360º
⇒ 90º + BOC + 90º + 40º = 360º
⇒ BOC = 360º – 220º = 140º

2. 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 3 cm (d) 4 3 cm

Ans.

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3. In figure if O is 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°

Ans. (a) 100°

OP ⊥ PR [∵ Tangent and radius are ⊥ to each other at the point of contact]
OPQ = 90º – 50º = 40º
OP = OQ [Radii]
∴ OPQ = OQP = 40º
In ΔOPQ,
⇒ POQ + OPQ + OQP = 180º
⇒ POQ + 40º + 40º = 180º
POQ = 180° – 80° = 100º.

4. In figure, O is the centre of a circle, AB is a chord and AT is the tangent at A. If AOB = 100°,
then BAT is equal to

(a) 100° (b) 40° (c) 50° (d) 90°

Ans. (c) 50°
AOB = 100°
OAB = OBA (∵ OA and OB are radii)
Now, in ∆AOB, AOB + OAB + OBA = 180° (Angle sum property of ∆)
⇒ 100° + x + x = 180°
[Let OAB = OBA = x]
⇒ 2x = 180° – 100° ⇒ 2x = 80° ⇒ x = 40°
Also, OAB + BAT = 90° [∵ OA is radius and TA is tangent at A]
⇒ 40° + BAT = 90° ⇒ BAT = 50°

5. In figure, PQ and PR are tangents to a circle with centre A. If QPA = 27°, then QAR equals to

(a) 63° (b) 153° (c) 126° (d) 117°

Ans. (c) 126°

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 QPA = RPA [∵ ∆AQP ≅ ∆ARP (RHS congruence rule)]
⇒ RPA = 27°
∴ QPR = QPA + RPA = 27° + 27° = 54°
Now, QAR + AQP + ARP + QPR = 360°
⇒ QAR = 90° + 90° + 54° = 360°
⇒ QAR = 360° – 234° = 126°

6. At one end A of 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) 6 cm (c) 8 cm (d) 5 cm
Ans. (c) 8 cm

7. In figure, AP, AQ and BC are tangents to the circle. If AB = 5 cm, AC = 6 cm and BC = 4 cm,
then the length of AP (in cm) is

(a) 7.5 (b) 15 (c) 10 (d) 9

Ans. (a) 7.5
AP = AQ

⇒ AB + BP = AC + CQ ⇒ 5 + BP = 6 + CQ
⇒ BP = 1 + CQ ⇒ BP = 1 + CR (∵ CQ = CR)
⇒ BP = 1 + (BC – BR)
⇒ BP = 1 + (4 – BP) (∵ BR = BP)
5
⇒ 2BP = 5 ⇒ BP = = 2.5 cm
2
Now, AP = AB + BP = 5 + 2.5 = 7.5 cm

8. In the figure PA and PB are tangents to the circle with centre O. If APB = 60°, then OAB is

(a) 30° (b) 60° (c) 90° (d) 15°

Ans. (a) 30°
Given APB = 60°
∵ APB + PAB + PBA = 180°
⇒ APB + x + x = 180° [∵ PA = PB ∴ PAB = PBA = x (say)]

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 ⇒ 60° + 2x = 180°
⇒ 2x = 180° – 60° ⇒ 2x = 120° ⇒ x = 60°
Also, OAP = 90°
⇒ OAB + PAB = 90°
⇒ OAB + 60° = 90° ⇒ OAB = 30°

In the following questions 9 and 10, a statement of assertion (A) is followed by a statement of
reason (R). Mark the correct choice as:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

9. Assertion (A): If the angle between two tangents drawn from an external point P to a circle of
radius 5 cm and centre O is 90°, then length of each tangent is 10 cm.
Reason (R): Opposite angles of a cyclic quadrilateral an supplementary.
Ans. (d) Assertion (A) is false but reason (R) is true.

10. Assertion (A): If the radius of a circle is 5 cm and distance of a point outside the circle from its
centre is 13 cm, then the length of the tangent drawn from that external point to the circle is 12
cm.
Reason (R): In a circle, tangent is always perpendicular to its radius at the point of contact.
Ans. (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of
assertion (A).

SECTION – B
Questions 11 to 14 carry 2 marks each.
11. In the given figure, PA and PB are tangents to the circle. CE is a tangent to the circle at D. If AP
= 15 cm, find the perimeter of the triangle PEC.

Ans. PA = PB = 15 cm (Tangent from P)

Perimeter of ∆PEC = PE + EC + PC
= PE + ED + DC + PC
= PE + EA + CB + PC [ED = EA and DC = CB]
= PA + PB = 15 + 15 = 30 cm

12. The radii of two concentric circles are 3 cm and 5 cm. Find the length of the chord of the outer
circle which is tangent to the smaller circle.
Ans. Radius is perpendicular to the tangent at the point of contact.

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 ∴ (OA)2 = (AM)2 + (OM)2 [Using Pythagoras Theorem]
2 2 2
⇒ (5) = (AM) + (3)
⇒ 25 = (AM)2 + 9
⇒ (AM)2 = 25 – 9
⇒ (AM)2 = 16
∴ AM = 4 cm
But perpendicular from centre bisects the chord
∴ AB = 2AM = 2 × 4 = 8 cm

13. In the given figure, ∆PQR is circumscribed, find x.

Ans. Tangents from an external point have equal length.

∴ PN = PM = 4 cm
⇒ MR = 12 – 4 = 8 cm
QL = QN = 6 cm
and RL = RM = 8 cm
⇒ x = QL + RL = 6 cm + 8 cm = 14 cm

14. In the given figure, O is the centre of the circle, PA and PB are tangent segments. Show that the
quadrilateral AOBP is a cyclic quadrilateral.

Ans. Radius is perpendicular to the tangent at the point of contact.

∴ ∠A = 90° and ∠B = 90°
In quadrilateral APBO,
∠A + ∠P +∠ B + ∠AOB = 360° [Angle sum property of quadrilateral]
⇒ 90° + ∠P + 90° + ∠AOB = 360°
∠P + ∠AOB + 180° = 360°
∠P + ∠AOB = 180°
But these are opposite angles of quadrilateral APBO
∴ AOBP is a cyclic quadrilateral.

SECTION – C
Questions 15 to 17 carry 3 marks each.

15. ABC is a right-angled triangle in which ∠B = 90° with AB = 6 cm and BC = 8 cm. A circle with
centre O has been inscribed inside the triangle. Find the value of x.

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 Ans. Method I:
Using Pythagoras Theorem,
AC2 = AB2 + BC2 = 62 + 82 = 100
⇒ AC = 10 cm
Join OA, OB and OC.
Then, ar(∆AOB) + ar(∆BOC) + ar(∆AOC) = ar(∆ABC)
1 1 1 1
⇒ × 6x + × 8x + × 10x = (6 × 8) ⇒ x = 2 cm
2 2 2 2
(As OR, OP and OQ are respectively perpendicular to BC, AB and AC. ∴ OR, OP and OQ are
altitudes)
Method II:
Each angle of quadrilateral BROP = 90° and OR = OP
⇒ BROP is a square.
∴ BR = x ⇒ CR = (8 – x) cm
and BP = x ⇒ AP = (6 – x) cm
Also, CQ = CR (Tangents from external point are equal)
and AP = AQ (Tangents from external point are equal)
AQ + CQ = 10 cm
⇒ 6 – x + 8 – x = 10 cm
⇒ x = 2 cm

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

Ans. Consider a parallelogram ABCD circumscribing a circle such that if touches the sides AB,
BC, CD and DA at P, Q, R and S respectively.

Now, we know lengths of tangents drawn from an external point to a circle are equal.
∴ AP = AS, BP = BQ, CR = CQ and DR = DS
Adding the above equations, we get
⇒ AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + BP) + (CR + DR) = (AS +DS) + (BQ + CQ)
⇒ AB + CD = AD + CB
But AB = CD and AD = CB [Since, opposite sides of parallelogram are equal]
⇒ AB + AB = AD + AD
⇒ 2AB = 2AD ⇒ AB = AD
⇒ AB = BC = CD = AD
Hence, ABCD is a rhombus.

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17. Prove that the intercepts of a tangent between two parallel tangents to a circle subtends a right
angle at the centre.
Ans: In the below figure, Join OC. Since, the tangents drawn to a circle from an external point
are equal.
∴ AP = AC

In Δ PAO and Δ AOC, we have:

AO = AO [Common]
OP = OC [Radii of the same circle]
AP = AC
⇒ Δ PAO ≅ Δ AOC [SSS Congruency]
∴ ∠PAO = ∠CAO = ∠1
∠PAC = 2 ∠1 ...(1)
Similarly ∠CBQ = 2 ∠2 ...(2)
Again, we know that sum of internal angles on the same side of a transversal is 180°.
∴ ∠PAC + ∠CBQ = 180°
⇒ 2 ∠1 + 2 ∠2 = 180° [From (1) and (2)]
⇒ ∠1 + ∠2 = 180°/2 = 90° ...(3)
Also ∠1 + ∠2 + ∠AOB = 180° [Sum of angles of a triangle]
⇒ 90° + ∠AOB = 180°
⇒ ∠AOB = 180° − 90° ⇒ ∠AOB = 90°.

SECTION – D
Questions 18 carry 5 marks.

18. (a) Prove that the tangent to a circle is perpendicular to the radius through the point of contact. (3)
(b) The length of the tangent to a circle of radius 3 cm is 4 cm from a point P. Find the distance
of P from the centre of the circle. (2)
Ans. (a) Given, To prove, Construction and figure of 1½ marks
Proof of 1½ marks
(b) Radius is perpendicular to the tangent at the point of contact.

∴ (OP)2 = (OA)2 + (PA)2 [Using Pythagoras Theorem]

= (3)2 + (4)2 = 9 + 16
2
(OP) = 25
∴ OP = 5 cm

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 SECTION – E (Case Study Based Questions)
Questions 19 to 20 carry 4 marks each.

19. Case Study -1:

An ambitious surveyor, tasked with designing a new park, discovered a triangular plot of land.
To maximize the usable space, they planned a perfect circular garden at its center. This circular
garden, with its center at point O, touches each of the three straight boundary fences, with the
line from O to D being perfectly perpendicular to the base fence BC.

Answer the questions based on above

(a) What will be the radius of the circle, if BD = 24 cm and OB = 25 cm? (1)
(b) Determine CD, if OC = 26 cm. (1)
(c) As AB and AC act as tangents to the circle at E and F and AE = 8 cm, then what is the
perimeter of ∆ABC. (2)
Ans. (a) In ∆OBD, OD ⊥ BD

(Radius is perpendicular to tangent)

OB2 = BD2 + OD2
⇒ (25)2 = (24)2 + (OD)2
⇒ 525 = 576 + OD2
⇒ OD2 = 625 – 576 = 49 ⇒ OD = 7 cm
Radius of circle = 7 cm
(b) In ∆OCD, OD ⊥ CD (Radius is perpendicular to tangent)

∴ OC2 = OD2 + CD2

⇒ (26)2 = (7)2 + CD2
⇒ 676 = 49 + CD2
⇒ CD2 = 676 – 49 = 627 cm
⇒ CD = 25.04 cm
(c)

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 As BD = BE
CD = CF ⇒ BF = 24 cm
CF = 25.0 cm
AE = AF = 8 cm
Perimeter of ∆ABC = AB + BC + AC
= (AE + BE) + (BD + CD) + (AF + FC)
= (8 + 24) + (24 + 25.04) + (8 + 25.04)
= 32 + 49.04 + 33.04 = 114.08 cm

20. Case Study – 2:

A community of birds, seeking a safe place to drink and bathe, banded together to build a
circular watering hole. They constructed it within the confines of a quadrilateral-shaped area of
land, with each of the four sides of their property acting as a perfect boundary for the new tank.
The sides AB, BC, CD, and DA measured 5m, 7m, 6m, and an unknown length respectively,
each serving as a tangent to their communal bird bath.

Answer the questions based on above.

(a) Find AD. (1)
(b) If O is centre of tank and AH & AE inclined to each other at angle 100°, then find ∠HOE. (1)
(c) If ∠GOF = (3x – 8)° and ∠GCF = (2x + 3)° then find the value of x. (1)
(d) Find the length of the tangent drawn from a point whose distance from the centre of a circle
is 25 cm and the radius of the circle is 7 cm. (1)
Ans. (a) AB + CD = AD + BC
⇒ 5 + 6 = AD + 7 ⇒ AD + 7 = 11 ⇒ AD = 4 cm
(b) ∠HOE + ∠HAE = 1800
⇒ ∠HOE + 1000 = 1800
⇒ ∠HOE = 800
(c) ∠GOF + ∠GCF = 1800 ⇒ 3x – 8 + 2x + 3 = 1800
⇒ 5x – 5 = 1800
⇒ 5x = 1800 + 50 = 1850
⇒ x = 370
(d) Let O is the centre of the circle and P is a point such that OP = 25 cm and PQ is the tangent
to the circle.

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 OQ = radius = 7 cm
In ∆OQP, we have ∠Q = 90°
OP2 = OQ2 + PQ2
⇒ (25)2 = 72 + PQ2
⇒ PQ2 = 625 – 49 = 576 ⇒ PQ = 24 cm
Hence, the length of the tangent = 24 cm

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