Please refer to Circle Class 9 Mathematics Exam Questions provided below. These questions and answers for Class 9 Mathematics have been designed based on the past trend of questions and important topics in your class 9 Mathematics books. You should go through all Class 9 Mathematics Important Questions provided by our teachers which will help you to get more marks in upcoming exams.

**Class 9 Mathematics Exam Questions Circle**

Class 9 Mathematics students should read and understand the important questions and answers provided below for Circle which will help them to understand all important and difficult topics.

**Very Short Answer Type Questions:**

**Question. In the given figure, O is the centre of circle. If ∠QPR = x, OM bisects ∠QOR and ∠QOM = y, then find the relation between x and y. **

**Ans.** Since OM bisects ∠QOR

∴ ∠QOR = 2 × ∠QOM = 2y …(i)

Since, angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

∴ 2∠QPR = ∠QOR

⇒ 2x = 2y [From (i)]

⇒ x = y

**Question. In the given figure, ABCD is a cyclic quadrilateral and O is the centre of the circle. If ∠BOD = 160°, then find the measure of ∠BPD. **

**Ans.** ∠BAD = 1/2 ∠BOD = 1/2 × 160° = 80°

Now, ∠BCD + ∠BAD = 180° (Opposite angles of a cyclic quadrilateral are equal)

⇒ ∠BCD + 80° = 180°

⇒ ∠BCD = 100°

Now, ∠BPD = ∠BCD = 100° (Angles in same segment are equal)

**Question. In the given figure, O is the centre of the circle. The angle subtended by the arc BCD at the centre is 140°. BC is produced to P. Determine ∠BAD. **

**Ans.** Since the angle subtended by an arc BCD at the centre is double the angle subtended by it at any point on the remaining part of the circle.

∴ ∠BAD = 1/2 ∠BOD

= 1/2 × (140°)= 70°

**Question. If ABCD is a cyclic quadrilateral in which AD || BC, then prove that ∠B = ∠C. ****Ans.** We have, ABCD is a cyclic quadrilateral and AD || BC.

∴ ∠A + ∠B = 180° …(i) Co-interior angles)

Also, ∠A + ∠C = 180° …(ii) (Opposite angles of a cyclic quadrilateral)

From (i) and (ii), we get ∠A + ∠B = ∠A + ∠C

∴ ∠B = ∠C

**Question. In the given figure, O is the centre of the circle. ∠CAB = 40°, ∠CBA = 110°, then find value of x. **

**Ans.** In ΔABC, 40° + 110° + ∠C = 180°

⇒ ∠C = 180° – 150° = 30°

Now, ∠AOB = 2 × ∠C

⇒ x = 2 × 30° = 60°.

**Short Answer Type Questions:**

**Question. In the given figure, AB and CD are two parallel chords of a circle with centre O and radius 13 cm such that AB = 10 cm and CD = 24 cm. If OP ⊥ AB and OQ ⊥ CD, find the length of PQ. **

**Ans.** Since OP ⊥ AB, OQ ⊥ CD and AB || CD, the points O, Q, P are collinear.

We know that the perpendicular from the centre of a circle to a chord bisects the chord.

∴ AP =1/2 AB = (1/2 x 10) cm = 5 cm

and CQ = 1/2CD = 1/2 (24) =12 cm

Join OA and OC.

Then, OA = 13 cm. [Radius of circle]

From the right angled ΔOPA,

we have

OP^{2} = (OA^{2} – AP^{2}) = [(13)^{2} – (5)^{2}] = 144

⇒ OP = 12 cm

Now, from right angled ΔOQC, we have

(OQ)^{2} = (OC)^{2} – (CQ)^{2} = 13^{2} – 12^{2} = 25

⇒ OQ = 5 cm

∴ PQ = OP – OQ = 12 – 5 = 7 cm

**Question. In the given figure, find out the value of (x – y) when ∠A = (2x + 4)°, ∠B = (x + 10)°, ∠C = (4y – 4)° and ∠D = (5y + 5)°. **

**Ans.** We know that the opposite angles of a cyclic quadrilateral are supplementary.

∴ ∠A + ∠C = 180° and ∠B + ∠D = 180°

⇒ 2x + 4 + 4y – 4 = 180

⇒ 2x + 4y = 180 ⇒ x + 2y = 90 …(i)

And x + 10 + 5y + 5 = 180

⇒ x + 5y = 165 …(ii)

Subtracting (i) from (ii), we get

3y = 75 ⇒ y = 25

Putting the value of y in (i), we get

x = 40

∴ x – y = 15

**Question. In the given figure, A, B, C are three points on a circle such that the angles subtended by the chords AB and AC at the centre O are 80° and 120° respectively. If ∠BAC is k°, then find the value of k. **

**Ans. ** Since arc BC makes ∠BOC at the centre and ∠BAC at a point on the remaining part of the circle.

∴ ∠BAC = 1/2 ∠BOC

Now, ∠BOC = 360° – (120° + 80°) = 160°

⇒ ∠BAC = 1/2 × 160° = 80° ∴ k = 80

**Question. AD is a diameter of a circle and AB is a chord. If AD = 58 cm, AB = 40 cm, then find the distance of AB (in cm) from the centre of the circle. **

**Ans.** The perpendicular drawn from centre to the chord bisects it.

∴ AM =1/2 AB =1/2 × 40 cm = 20 cm

Also, OA = 1/2 AD = 1/2 × 58 cm = 29 cm

In right ΔOAM, we have OA^{2} = OM^{2} + AM^{2}

⇒ 29^{2} = OM^{2} + 20^{2} ⇒ 841 = OM^{2} + 400

⇒ OM^{2 } = 841 – 400 ⇒ OM^{2 }= 441 ⇒ OM = 21 cm.

**Question. In the given figure, if x/3 = y/4 = z/5 , then calculate the values of x, y and z. **

**Ans.** Let x/3 = y/4 = z/5 =k

Then, x = 3k, y = 4k and z = 5k

∠BCP = ∠DCQ = 3k

[Vertically opposite angles]

In ΔDCQ,

∠CDQ = 180° – (3k + 5k) = 180° – 8k

∠ADC + ∠CBA = 180°

(Opposite angles of cyclic quadrilateral)

⇒ 180° – (180° – 8k) + ∠CBA = 180°

⇒ ∠CBA = 180° – 8k

∴ ∠PBC = 180° – ∠CBA = 180° – (180° – 8k) = 8k

In ΔPBC, ∠P + ∠B + ∠C = 180°

⇒ 4k + 8k + 3k = 180° ⇒ 15k = 180°

⇒ k = 12 ∴ x = 36°, y = 48° and z = 60°

**Question. Prove that, among any two chords of a circle, the larger chord is nearer to the centre.****Ans.** Let AB be the larger chord and CD be the shorter one.

Draw OM ⊥ AB

⇒ AM =1/2 AB

Draw ON ⊥ CD

⇒ CN = 1/2 CD

In ΔOAM and ΔOCN

OA^{2} = OM^{2} + AM^{2} and OC^{2} = ON^{2} + CN^{2}

But OA = OC ⇒ OA^{2} = OC^{2} (Radii of same circle)

⇒ OM^{2} + AM^{2} = ON^{2} + CN^{2}

⇒ AM^{2} – CN^{2} = ON^{2} – OM^{2} …(i)

Since AB > CD ⇒ 1/2 AB >1/2 CD ⇒ AM > CN

⇒ AM^{2} > CN^{2} ⇒ AM^{2} – CN^{2} > 0

⇒ ON^{2} – OM^{2} > 0 [Using (i)]

⇒ ON^{2} > OM^{2} ⇒ ON > OM or OM < ON

Thus, AB is nearer to the centre.

**Question. In the given figure, ABCD is a cyclic quadrilateral. Find the value of x/50°.**

**Ans.** We have, ∠CDA = 180° – 80° = 100°

and ∠ABC + x = 180°

Now, ∠CDA + ∠ABC = 180°

⇒ 100° + 180° – x = 180°

⇒ x = 100°

∴ x / 50 = 100° / 50° = 2

**Question. In the given figure, O is the centre and AE is the diameter of the semi circle ABCDE. If AB = BC and ∠AEC = 45°, then find (i) ∠CBE (ii) ∠CDE (iii) ∠AOB. And prove that BO || CE. **

**Ans. **Join OC.

∠AOC = 2∠AEC ⇒ ∠AOC = 2 × 45° = 90°

In ΔAOB and ΔBOC, AB = BC, AO = OC and OB = OB

∴ ΔBOA ≅ ΔBOC (By SSS congruency)

∴ ∠BOA = ∠BOC (CPCT)

∠BOA + ∠BOC = ∠AOC = 90°

⇒ 2∠BOA = 90°

⇒ ∠BOA = 45° and ∠BOC = 45°

⇒ ∠BOA = ∠CEO

(Each equal 45°)

But they are corresponding angles.

∴ BO || CE

Now, ∠AOC + ∠COE = 180°

∴ ∠COE = 90° ( ∠AOC = 90°)

Now, 2∠CBE = ∠COE = 90°

∴ ∠CBE = 45°

Since BCDE is a cyclic quadrilateral.

∴ ∠CBE + ∠CDE = 180° ⇒ 45° + ∠CDE = 180°

⇒ ∠CDE = 180° – 45° = 135°

**Long Answer Type Questions:**

**Question. PQ and RS are two parallel chords of a circle whose centre is O and radius is 10 cm. If PQ = 16 cm and RS = 12 cm, find the distance between PQ and RS, if they lie****(i) on the same side of the centre O.****(ii) on the opposite side of the centre O.****Ans. ** We have, OP = OR = 10 cm, PQ = 16 cm and RS = 12 cm Now, draw a perpendicular from centre O to chords PQ and RS, meets at L and M respectively i.e., OL ⊥ PQ and OM ⊥ RS.

∴ PL = 8 cm and RM = 6 cm

[∴ Perpendicular drawn from the centre of a circle to a chord bisects the chord]

In right triangles OLP and OMR, we have

OP^{2} = OL^{2} + PL^{2} and OR^{2} = OM^{2} + RM^{2}

[By Pythagoras theorem]

⇒ 100 = OL2 + 64 and 100 = OM2 + 36

⇒ OL^{2} = 36 and OM^{2} = 64

⇒ OL = 6 cm and OM = 8 cm

(i) In this case, from fig. I, we have

Distance between PQ and RS

= LM = OM – OL = (8 – 6) cm = 2 cm

(ii) In this case, from fig. II, we have Distance between PQ and RS = LM

= OL + OM = (6 + 8) cm = 14 cm

**Question. PQ and PR are the two chords of a circle of radius r. If the perpendiculars drawn from the centre of the circle to these chords are of lengths a and b, respectively and PQ = 2PR, then prove that b2 = a ^{2}/4 + 3/4r^{2}Ans.** Join OP.

Since the perpendicular from the centre of the circle to the chord bisects the chord.

∴ PM = MQ =1/2 PQ and PL = LR = 1/2 PR

In right angled ΔOMP, we have

PM^{2} = OP^{2} – OM^{2}

{1/2 PQ}^{2} = r^{2} – a^{2} ⇒ PQ^{2}/4 = r^{2} – a^{2}

⇒ PQ^{2} = 4r^{2} – 4a^{2} …(i)

Again, in right angled ΔOLP, we have

PL^{2} = OP^{2} – OL^{2}

⇒ {1/2 PR}^{2} = r^{2} – a^{2} ⇒ PR2 = 4r^{2} – 4b^{2 }…(ii)

Also, PQ = 2PR [Given]

⇒ PQ2 = 4PR2 …(iii)

From (i) , (ii) and (iii), we have

4r^{2} – 4a^{2} = 4(4r^{2} – 4b^{2}) ⇒ r^{2} – a^{2} = 4r^{2} – 4b^{2}

⇒ 4b^{2} = 4r^{2} – r^{2} + a^{2} ⇒ 4b^{2} = 3r^{2} + a^{2}

⇒ b^{2} = 3/4 r^{2} + 1/4 a^{2}

or b^{2} = a^{2}/4 + 3/4 r^{2}

**Question. In the adjoining figure, O is the centre of a circle. If AB and AC are chords of the circle such that AB = AC, OP ⊥ AB and OQ ⊥ AC, then prove that PB = QC.**

**Ans. **Given : AB and AC be two equal chords of circle with centre O. Also, OP ⊥ AB at M and OQ ⊥ AC at N.

To prove : PB = QC.

Proof : We know that, the perpendicular drawn from the centre of a circle to a chord bisects the chord.

∴ AM = MB = 1/2 AB [∴ OP ⊥ AB]

and AN = NC = 1/2 AC [∴ OQ ⊥ AC]

Since, AB = AC [Given]

∴ 1/2 AB = 1/2 AC

⇒ AM = MB = AN = NC …(i)

⇒ OM = ON …(ii)

[∴ Equal chords of a circle are equidistant from the centre]

⇒ OP = OQ [Radii of same circle] …(iii)

⇒ OP – OM = OQ – ON [Subtracting (ii) from (iii)]

⇒ PM = QN

Now, in ΔPMB and ΔQNC, we have

MB = NC [From (i)]

∠PMB = ∠QNC [Each equal to 90°]

PM = QN [Proved above]

∴ ΔPMB ≅ ΔQNC [By SAS congruency criteria]

⇒ PB = QC [By C.P.C.T]