## Download PDF of ICSE Class 10 Circles Previous Years’ Questions

## ICSE Class 10 Circles Previous Years Questions + Solution

**1. In the given figure, O is the centre of the circle. PQ is a tangent to the circle at T. Chord AB produced meets the tangent at P. AB = 9 cm, BP = 16 cm, ∠PTB = 50°, ∠OBA = 45°. Find:**

**(a) length of PT(b) ∠BAT(c) ∠BOT(d) ∠ABT [2023]**

**Answer: (a) 20cm (b) 50° (c) 100° (d) 85°**

**Step-by-step Explanation:**

(a) We know, PT^{2} = AP x BP (When tangent and chord intersect externally, the product of the lengths of the segments of chord is equal to the square of the length of the tangent.)

PT^{2} =(16+9) x 16

PT^{2} =25 x 16

PT = √25×16

PT= 20 cm

(b) ∠BAT= ∠BTP = 50° (angle in the alternate segment)

(c) ∠BOT =2∠BAT= 100° (Angle subtended by an arc at the center of a circle is double the angle subtended by it on remaining part of the circle.)

(d) In ΔBOT, OB = OT (radii of a circle)

∴∠OBT=∠BTO= 180°-100°/2= 40°

∴∠ABT= 45° + 40°= 85°

**Get the video solution of ICSE Class 10 Circles Previous Years Questions HERE. **

**2. In the given diagram RT is a tangent touching the circle at S. If ∠PST= 50° and ∠SPQ = 60° then ∠PSQ is equal to:**

**(a) 40°(b) 30°(c) 60°(d) 90° [2023]**

**Answer: (d)**

**Step-by-step Explanation:**

∠PQS=∠PST= 30°

In ΔPQS, ∠PQS +∠PSQ+∠QPS= 180°

∠PSQ= 180° – (60+30)°= 90°

**3. In the given figure O, is the centre of the circle. CE is a tangent to the circle at A.If ∠ABD=26°, then find**

**(a) ∠BDA(b) ∠BAD(c) ∠CAD(d) ∠ODB [2023]**

**Answer:** (a) 90° (b) 64° (c) 26° (d) 26°

**Step-by-step Explanation:**

(a) **∠**BDA= 90° (angle in a semicircle is right angle.)

(b) **∠**BAD= 180° – (90+26)° (sum of angles of a triangle is 180°

= 64°

(c) **∠**CAD= 26° (angles in the alternate segments are equal.)

(d) **∠**DOB= 2**∠**BAD= 2× 64= 128° (angle subtended by an arc at the center of a circle is double the angle subtended by it on any part on the remaining circle.)

∴ ∠ODB= 180° – (26+128)° =26° (sum of the angles of a triangle is 180°.)

**4. In the given figure A, B, C and D are points on the circle with centre O. Given ∠ABC = 62°**

Find:

(a) ∠ADC

(b) ∠CAB `[2022 Semester-2]`

**Solution:** **(a) 62° (b) 28°**

**Step-by-step Explanation:**

(a) ∠ADC= ∠ABC= 62**°** (Angles in the same segment are equal.)

(b) ∠ACB= 90**°** (angle in a semicircle is right angle.)

∴ ∠CAB= 180**°**– (62**°**+90**°**) =28**°** (sum of angles in a triangle is 180**°**.)

**5. Two chords AB and CD of a circle intersect externally at E. If EC = 2 cm, EA = 3 cm and AB = 5 cm, Find the length of CD. [2022 Semester-2]**

**Answer: 10 cm**

**Step-by-step Explanation:**

We know, AE × BE = CE × DE (when two chords intersect internally or externally, the products of the lengths of the segments of the chords are equal.)

3 × (3+5) = 2 × (2+CD)

24/2 = 2 + CD

10 = CD

CD = 10 cm

**6. In the given figure O is the centre of the circle. PQ and PR are tangents and ∠QPR = 70°. Calculate**

**(a) ∠QOR(b) ∠QSR [2022 Semester-2]**

**Answer: (a) 110° (b) 125° **

**Step-by-step Explanation: **

(a) ∠PQO= ∠PRO = 90° (tangent and the radius of a circle through the point of contact are perpendicular to each other.)

In Quadrilateral PQOR,

∠RPQ +∠PQO +∠QOR + ∠PRO = 360°

70° + 90° +∠QOR + 90° = 360°

∠QOR = 360° – 250° = 110°

(b) reflex ∠QOR = 360°-110° = 250°

∠QSR = 125° (angle subtended by an arc at the center of a circle is double the angle subtended by it on any part on the remaining circle.)

**ICSE Class 10 Maths PYQs chapter-wise**

**7. ABCD is a cyclic quadrilateral. If ∠BAD = (2x + 5)° and ∠BCD = (x + 10)° then x is equal to:(a) 65° (b) 45° (c) 55° (d) 5° [2022 Semester-2]**

**Answer:** **(c)**

**Step-by-step Explanation:**

We know, by theorem, opposite angles of a cyclic quadrilateral are supplementary.

∴ ∠BAD + ∠BCD = 180°

(2x + 5)° + (x + 10)° = 180°

3x +15 = 180

3x = 165

x = 55°

**8. In the given figure TP and TQ are two tangents to the circle with centre O, touching at A and C, respectively. If ∠BCQ = 55° and ∠BAP = 60°, find:**

**(i) ∠OBA and ∠OBC(ii) ∠AOC(iii) ∠ATC [2020]**

**Answer: (i) 30°**,** 35° (ii) 130° (iii) 50° **

**Step-by-step Explanation:**

(i) PAT and QCT are tangents to the circle.

∴ ∠QCO = ∠PAO = 90° (tangent and the radius of a circle through the point of contact are perpendicular to each other.)

Now, ∠BCQ = 55°.

∴ ∠BCO = 90 – 55 = 35°

In ΔBOC, OB = OC (radii)

∴ ∠OBC = ∠OCB = 35°

Similarly,

∠BAO = 90 – 60 = 30°

In ΔOAB, OA = OB (radii)

∴ ∠OBA = ∠BAO = 30°

(ii) ∠ABC= ∠OBA + ∠OBC = 30 + 35 = 65°

Hence, ∠AOC= 2∠ABC =130° (angle subtended by an arc at the center is double the angle subtended by it on the remaining part on the circle.)

(iii) ∠ATC= 360° – (∠TAO + ∠AOC + ∠TCO) (sum of angles of a quadrilateral is 360°.)

∴ ∠ATC= 360° – (90+130+90)°= 360°-310°= 50°

**9. In the given figure AB = 9 cm, PA = 7.5 cm and PC = 5 cm. Chords AD and BC intersect at P.**

**(i) Prove that ΔPAB ~ ΔPCD(ii) Find the length of the CD.(iii) Find area of ΔPAB : area of ΔPCD [2020]**

**Answer: (ii) 6 cm (iii) 9 : 4 **

**Step-by-step Explanation:**

(i) Chords AD and BC intersect internally. Therefore according to the theorem, the product of the lengths of their segments are equal.

∴ AP × PD = BP × PC

or, AP/PC = BP/PD

Now, In ΔPAB and ΔPCD

∠APB= ∠CPD (vertically opposite angles)

AP/PC = BP/PD (proved above)

∴ ΔPAB ~ ΔPCD (S-A-S condition of similarity)

(ii) As ΔPAB ~ ΔPCD

∴ AP/PC = BP/PD = AB/CD

AP/PC = AB/CD

7.5/5 = 9/CD

CD= 9/1.5= 6 cm

(iii) Area of ΔPAB : Area of ΔPCD = (PA/PC)^{2} (ratio of areas of similar triangles is equal to the square of the ratio of their corresponding sides.)

Area of ΔPAB : Area of ΔPCD = (7.5/5)^{2} =9 : 4

**10. In the figure given below, O is the centre of the circle and AB is a diameter.**

**If AC = BD and ∠AOC = 72°. Find:(i) ∠ABC(ii) ∠BAD(iii) ∠ABD [2020]**

**Answer:** **(i) 36° (ii) 36° (iii) 54°**

**Step-by-step Explanation:**

(i) ∠ABC = 1/2∠AOC =36° (angle subtended by an arc at the center is double the angle subtended by it on the remaining part on the circle.)

(ii) ∠BAD= ∠ABC = 36° (equal chords subtend equal angles.)

(iii) ∠ADB= 90° (angle in a semicircle is right angle.)

∴ ∠ABD= 180°- (∠BAD + ∠ADB) (sum of angles of a triangle is 180°.)

or, ∠ABD= 180° – 126°= 54°

**11. In the given figure, AC is a tangent to the circle with center 0. If ∠ADB = 55°, find x and y. Give reasons for your answers. [3] [2019]**

**Answer:** **x= 20° , y= 70°**

**Step-by-step Explanation:**

∠AEB= 90° (angle in a semicircle is right angle.)

∴ ∠AED= 90° (linear pair)

∠DAE= 180°-(90°+55°)= 35°

∴∠ABE= 35° (angles in the alternate segments are equal.)

∴ ∠AOE= y°= 70° (angle subtended by an arc at the center is double the angle subtended by it on the remaining part on the circle.)

∠OEB=∠OBE= 35° (isosceles triangle property)

Hence, ∠DEC=∠OEB= 35°

∠EDC= 180 – 55=125° (linear pair)

Hence, x°= 180° – (125 + 35)° = 20°

**12. In the given figure, ABCDE is a pentagon inscribed in a circle such that AC is a diameter and side BC || AE. If ∆BAC = 50°, find giving reasons : [4]**

**(i) ∠ACB(ii) ∠EDC(iii) ∠BEC [2019]**

**Answer:** **(i) 40° (ii) 140° (iii) 50°**

**Step-by-step Explanation:**

(i) ∠ABC= 90° (angle in a semicircle is right angle.)

Hence, ∠ACB= 180° – (90+50)°= 40°

(ii) ∠CAE=∠ACB= 40°

Hence, ∠EDC= 180° – 40°= 140° (opposite angles of a cyclic quadrilateral are supplementary.)

(iii) ∠BEC= ∠BAC= 50° (angles in the same segment are equal.)

**13. PQRS is a cyclic quadrilateral. Given ∠QPS = 73°, ∠PQS = 55° and ∠PSR = 82°, calculate: [4]**

**(i) ∠QRS(ii) ∠RQS(iii) ∠PRQ [2018]**

**Answer: (i) 107° (ii) 43° (iii) 52° **

**Step-by-step Explanation:**

(i)** **∠QRS= 180°- 73° = 107° (opposite angles of a cyclic quadrilateral are supplementary.)

(ii) ∠PSQ= 180° – (73 +55)°=52°

∴ ∠RSQ= 82 – 52= 30°

Hence, ∠RQS= 180°- (107 + 30)°= 43°

(iii) ∠PRQ= ∠PSQ= 52° (angles in the same segment are equal.)

**14. In the figure given below ‘O’ is the center of the circle. If QR = OP and ∠ORP= 20°. Find the value of ‘x ’ giving reasons. [3] [2018]**

**Answer:** **60°**

**Step-by-step Explanation:**

OP=QR (given) and OP= OQ (radii)

Hence, OQ = QR

∴ ∠QOR =∠ORQ= 20°

∴ ∠OQR= 180° – 40°= 140° (angle sum property of triangle)

∴ ∠OQP= 180° – 140° =40°

∴ ∠OPQ= 40°

∴ ∠POQ= 180° – 80° = 100°

∴ x°= 180°- (100 +20)° = 60° (angles in a straight line)

**15. AB and CD are two parallel chords of a circle such that AB = 24 cm and CD = 10 cm. If the radius of the circle is 13 cm, find the distance between the two chords. [3] [2017]**

**Answer:** **17 cm**

**Step-by-step Explanation:**

Join OB and OD,

NB= 1/2 AB= 12 cm and MD= 1/2 CD= 5 cm (perpendicular drawn from the center of a circle to the chord bisects it.)

In ΔONB, By pythagoras theorem,

ON = √OB^{2} -NB^{2}

ON = √169 – 144 = 5 cm

In ΔOMD, By pythagoras theorem,

OM = √OD^{2} – MD^{2}

ON = √169 – 25 = 12 cm

∴ MN= 5 + 12= 17 cm

**16. In the given figure PQ is a tangent to the circle at A. AB and AD are bisectors of ∠CAQ and ∠PAC. If ∠BAQ = 30° prove that :**

(i) BD is a diameter of the circle.

(ii) ABC is an isosceles triangle. [2017]

**Step-by-step Explanation: **

(i) Given that AB and AD are bisectors of ∠CAQ and ∠PAC.

Let ∠CAB= ∠BAQ= x° and ∠CAD= ∠DAP= y°.

∴ ∠BAQ +∠CAB +∠CAD +∠DAP= (2x + 2y)°

(2x + 2y)°= 180° (angles in a straight line.)

2(x+y) = 180°

x + y = 90°

or, ∠BAD = 90°

Hence, BD is the diameter of the circle. (angle in a semicircle is right angle.)

(ii) ∠ACB = ∠BAQ = x° (angles in the alternate segments are equal.)

∠CAB = x°

∴ AB = BC

Hence, ABC is an isosceles triangle.

**Get the video solution of ICSE Class 10 Circles Previous Years Questions HERE. **

**17. In the figure given, O is the center of the circle. ∠DAE = 70°. Find giving suitable reasons, the measure of: [4]**

(i) ∠BCD

(ii) ∠BOD

(iii) ∠OBD `[2017]`

**Answer: (i) 70° (ii) 140° (iii) 20°**

**Step-by-step Explanation: **

(i) ∠BCD = ∠DAE = 70**°** (exterior angle of a cyclic quadrilateral is equal to opposite interior angle.)

(ii) ∠BOD = 2∠BCD = 140**°** (angle subtended by an arc at the center is double the angle subtended by it on the remaining part on the circle.)

(iii) In ΔBOD, OB = OD (radii)

∴ ∠OBD = ∠ODB = 180 – ∠BOD / 2 = 20**°**

**18. In the given figure below, AD is a diameter. O is the centre of the circle. AD is parallel to BC and ∠CBD = 32°.**

Find :

(i) ∠OBD

(ii) ∠AOB

(iii) ∠BED [4] [2016]

**Answer: (i) 32°** **(ii) 64° (iii) 58°**

**Step-by-step Explanation:**

(i) Since AD is parallel to BC,

∠ODB = ∠CBD = 32**°** (alternate interior angles)

∠OBD =∠ODB = 32**°** (property of isosceles triangle.)

(ii) ∠AOB = 2∠ODB = 64**°** (angle subtended by an arc at the center is double the angle subtended by it on the remaining part on the circle.)

(iii) ∠BOD = 180**°** – ∠AOB = 180**°** – 64**°** = 116**°**

Hence, ∠BED = 116/2 = 58**°** (angle subtended by an arc at the center is double the angle subtended by it on the remaining part on the circle.)

**19. In the figure given below, O is the centre of the circle and SP is a tangent. If ∠SRT = 65°, find the value of x, y and z. [4] [2015]**

**Answer: x=25°, y= 50°, z= 40°**

**Step-by-step Explanation:**

∠RST= 90**°** (tangent and the radius of a circle through the point of contact are perpendicular to each other.)

Hence, ∠RTS= x= 180**°** – (65 + 90)**°**= 25**°**

y = ∠SOQ= 2∠RST= 50° (angle subtended by an arc at the center is double the angle subtended by it on the remaining part on the circle.)

z= ∠OPS= 180° – (90 + 50)°= 40°

**20. AB and CD are two chords of a circle intersecting at P. Prove that AP × PB = CP × PD. [3] [2015]**

**Step-by-step Explanation:**

Let us join AD and BC.

Now, In ΔAPD and ΔCPB,

∠A =∠C (angles in the same segment are equal.)

∠APD = ∠BPC (vertically opposite angles)

∴ ΔAPD ~ ΔCPB (A-A condition of similarity)

∴ AP/CP = PD/PB

or, AP × PB = CP × PD

Proved.

**21. In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate: [3]**

(i) ∠BDC

(ii) ∠BEC

(iii) ∠BAC [2014]

**Answer: (i) 32° (ii) 148° (iii) 32° **

**Step-by-step Explanation:**

(i) ∠BCD= 90**°** (angle in a semicircle is right angle.)

∴ ∠BDC= 180**°**– (∠BCD + ∠DBC)

=180**°** – (90 + 58)**°**

= 180**°** – 148**°** = 32**°**

(ii) ∠BEC = 180**°** – ∠BDC = 180**°** – 32**°**= 148° (opposite angles of a cyclic quadrilateral are supplementary.)

(iii) ∠BAC = ∠BDC = 32**°** (angles in the same segment are equal.)

**22. In the figure given below, diameter AB and CD of a circle meet at P. PT is a tangent to the circle at T. CD=7.8 cm, PD=5 cm, PB=4 cm.**

**Find:(i) AB.(ii) the length of tangent PT. [3][2014]**

**Answer:** **(i) 12 cm (ii) 8 cm**

**Step-by-step Explanation:**

(i) We know, When two chords intersect internally or externally, the product of the lengths of the segments of the chords are equal.

∴ AP × PB = CP × PD

or, (AB +4) × 4 = (7.8 + 5) × 5

or, AB + 4 = 12.8 × 5/4

or, AB = 16 – 4 = 12 cm

(ii) We know, When a tangent and a chord of a circle intersect externally, the product of the lengths of the segments of the chord is equal to the square of the length of the tangent.

∴ PT^{2} = AP × PB

or, PT^{2} = 16× 4

or, PT = √64

PT = 8

**23. In the figure given below, O is the centre of the circle. AB and CD are two chords of the circle. OM is perpendicular to AB and ON is perpendicular to CD. AB = 24 cm, OM = 5 cm, ON = 12 cm. **

**Find the :(i) radius of the circle.(ii) length of chord CD. [3] [2014]**

**Answer:** **(i) 13 cm (ii) 10 cm**

**Step-by-step Explanation:**

Let us join CO and AO.

(i) In ΔAMO, AM = 24/2 = 12 cm. (Perpendicular drawn from the centre of a circle to the chord bisects it.)

∴ by pythagoras theorem,

AO = √AM^{2} + OM^{2}

AO = √12^{2} + 5^{2}

AO = √169 = 13 cm

∴ radius = 13 cm

(ii) In ΔCNO,

CN = √CO^{2} – ON^{2}

CN = √13^{2} – 12^{2}

CN = √25 = 5 cm

∴ CD = 2 CN = 10 cm

**24. In the given figure, **

**∠BAD = 65°, **

**∠ABD = 70°, **

**∠BDC = 45°**

i.)Prove that AC is a diameter of the circle.

ii.)Find ∠ACB.**[2013]**

**Answer:** **(ii) 45° **

**Step-by-step Explanation:**

(i)** **In ΔADB,

∠ADB = 180°- (65+70)°= 45°

∴∠ADC = (45+45)°= 90°

Hence, AC is the diameter of the circle. (angle in a semicircle is a right angle.)

(ii) ∠ACB =∠ADB = 45°

**25. In the given circle with centre O, ∠ABC = 100°, ∠ACD = 40° and CT is a tangent to the circle at C.Find ∠ADC and ∠DCT. [2013]**

**Answer:** **∠ADC = 80° and ∠DCT= 60° **

**Step-by-step Explanation:**

∠ADC= 180°-∠ABC (opposite angles of a cyclic quadrilateral are supplementary.)

∠ADC= (180-100)° = 80°

∠DAC= 180°-(40 + 80)° = 60°

∴∠DCT = ∠DAC= 60° (angles in the alternate segment are equal.)

**26. In the given figure O is the centre of the circle and AB is a tangent at B. If AB = 15 cm and AC = 7.5 cm. Calculate the radius of the circle. [3] [2012]**

**Answer:** **11.25 cm**

**Step-by-step Explanation:**

We know, When a tangent and a chord of a circle intersect externally, the product of the lengths of the segments of the chord is equal to the square of the length of the tangent.

∴ AB^{2} = AC × AD

or, 15^{2} = 7.5 × (CD + 7.5)

or, CD + 7.5 = 225/7.5

or, CD = 30 – 7.5 = 22.5 cm

CD is the diameter of the circle.

∴ radius= 22.5/2= 11.25 cm

**27. In the given figure AB is the diameter of a circle with centre O. ∠BCD = 130°. Find:**

(i) ∠DAB

(ii) ∠DBA. [3] [2012]

**Answer: (i) ∠DAB=50° (ii) ∠DBA= 40° **

**Step-by-step Explanation:**

(i) ∠DAB= 180°-130°= 50° (opposite angles of a cyclic quadrilateral are supplementary.)

(i) ∠ADB= 90° (angle in a semicircle is right angle.)

∴ In ΔADB,

∠DBA= 180°- (90+50)°=40°

**28. In triangle PQR, PQ = 24 cm, QR = 7 cm and ∠PQR = 90°. Find the radius of the inscribed circle. [2012]**

**Answer:** **3cm**

**Step-by-step Explantion:**

PR^{2} =QR^{2} + PQ^{2}

PR^{2} = 7^{2} + 24^{2}

PR= √625 = 25 cm

ON and OM are joined.

We know, tangent to a circle and radius through the point of contact are perpendicular to each other.

∴ ∠ONQ= 90° and ∠OMQ= 90°

QM = QN (tangents drwan from an external point to a circle are equal in length.)

∴ OMQN is a square with each side x cm.

∴ MR= TR=(7-x) cm, PN=PT= (24-x) cm

Now, PR= PT + TR

(24-x) + (7-x) = 25

or, 31 – 2x = 25

or, 2x = 6

or, x= 3 cm

Hence, radius= 3cm

**29. In the given figure O is the centre of the circle. Tangents at A and B meet at C. If ∠AOC = 30°, find**

**(i) ∠BCO(ii) ∠AOB(iii) ∠APB [3] [2011]**

**Answer:** **(i) 30° (ii) 120° (iii) 60°**

**Step-by-step Explanation:**

(i) ∠BCO = ∠ACO = 30° (two tangents drawn from an external point to a circle are equally inclined to the line segment joining the centre to that point.)

(ii) ∠OAC = ∠OBC = 90° (tangent to a circle and radius through the point of contact are perpendicular to each other.)

∴ ∠AOB = 360°- (90 +30 + 30 +90)° = 120°

(iii) ∠APB = 1/2 of ∠AOB = 1/2 of 120°= 60°

**30. ABC is a triangle with AB = 10 cm, BC = 8 cm and AC = 6 cm (not drawn to scale). Three circles are drawn touching each other with the vertices as their centres. Find the radii of the three circles. [3][2011]**

**Answer: 6 cm, 2 cm, 4 cm**

**Step-by-step Explanation:**

Let the radii of the three circles be r_{1} , r_{2} , and r_{3} respectively.

so, BC= r_{1} + r_{2} = 8 …… (1)

AC= r_{2} + r_{3} = 6 ….. (2)

AB= r_{1} + r_{3} = 10 ….. (3)

Adding (1), (2) and (3) we get,

2( r_{1} + r_{2} + r_{3})= 24

r_{1} + r_{2} + r_{3} =12 …. (4)

subtracting (1) from (4), we get,

(r_{1} + r_{2} + r_{3})-(r_{1} + r_{2})=12- 8

r_{3} =4 cm

Similarly, subtracting (2) from (4), we get,

r_{1} = 6 cm and

subtracting (3) from (4), we get,

r_{2} = 2 cm

**31. In the following figure O is the centre of the circle and AB is a tangent to it at point B. ∠BDC = 65°. Find ∠BAO. [3] [2010]**

**Answer:** **40°**

**Step-by-step Explanation:**

In ΔCBD, ∠ABO= 90° (tangent to a circle and radius through the point of contact are perpendicular to each other.)

∴ ∠BCD= 180°-(65 + 90)°= 25°

∴ ∠BOA= 2∠BCD= 50° (angle subtended by an arc at the centre of a circle is double the angle subtended by it on the remaining part of the circle.)

In ΔAOB, ∠BAO= 180°-(90 + 50)°= 40°

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