# Circle Class 9 Mathematics Exam Questions

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.

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.

= 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°.

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
OP2 = (OA2 – AP2) = [(13)2 – (5)2] = 144
⇒ OP = 12 cm
Now, from right angled ΔOQC, we have
(OQ)2 = (OC)2 – (CQ)2 = 132 – 122 = 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 OA2 = OM2 + AM2
⇒ 292 = OM2 + 202 ⇒ 841 = OM2 + 400
⇒ OM = 841 – 400 ⇒ OM= 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
⇒ 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
OA2 = OM2 + AM2 and OC2 = ON2 + CN2
But OA = OC ⇒ OA2 = OC2 (Radii of same circle)
⇒ OM2 + AM2 = ON2 + CN2
⇒ AM2 – CN2 = ON2 – OM2 …(i)
Since AB > CD ⇒ 1/2 AB >1/2 CD ⇒ AM > CN
⇒ AM2 > CN2 ⇒ AM2 – CN2 > 0
⇒ ON2 – OM2 > 0 [Using (i)]
⇒ ON2 > OM2 ⇒ 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°

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
OP2 = OL2 + PL2 and OR2 = OM2 + RM2
[By Pythagoras theorem]
⇒ 100 = OL2 + 64 and 100 = OM2 + 36
⇒ OL2 = 36 and OM2 = 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 = a2/4 + 3/4r2
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
PM2 = OP2 – OM2
{1/2 PQ}2 = r2 – a2 ⇒ PQ2/4 = r2 – a2
⇒ PQ2 = 4r2 – 4a2 …(i)
Again, in right angled ΔOLP, we have
PL2 = OP2 – OL2
⇒ {1/2 PR}2 = r2 – a2 ⇒ PR2 = 4r2 – 4b2 …(ii)
Also, PQ = 2PR [Given]
⇒ PQ2 = 4PR2 …(iii)
From (i) , (ii) and (iii), we have
4r2 – 4a2 = 4(4r2 – 4b2) ⇒ r2 – a2 = 4r2 – 4b2
⇒ 4b2 = 4r2 – r2 + a2 ⇒ 4b2 = 3r2 + a2
⇒ b2 = 3/4 r2 + 1/4 a2
or b2 = a2/4 + 3/4 r2

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]