Chapter 9 Circles NCERT Exemplar Solutions Exercise 9.1 Class 10 Maths
Chapter Name  NCERT Maths Exemplar Solutions for Chapter 9 Circles Exercise 9.1 
Book Name  NCERT Exemplar for Class 10 Maths 
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Related Study  NCERT Solutions for Class 10 Maths 
Exercise 9.1 Solutions
Multiple Choice Questions
Choose the correct answer from the given four options :
1. 1. If radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle which is tangent to the other circle is
(A) 3 cm
(B) 6 cm
(C) 9 cm
(D) 1 cm
Solution
As given in the question,
OA = 4cm,
OB = 5cm
And,
OA ⊥ BC
Therefore,
OB^{2} = OA^{2} + AB^{2}
⇒ 5^{2} = 4^{2} + AB^{2}
⇒ AB = 3cm
And,
BC = 2AB
= 2 × 3cm
= 6cm
(A) 62.5°
(B) 45°
(C) 35°
(D) 55°
Solution
ABCD is a quadrilateral circumscribing the circle
And we know that the opposite sides of a quadrilateral circumscribing a circle subtend
supplementary angles at the center of the circle.
So,
∠AOB + ∠COD = 180°
⇒ 125° + ∠COD = 180°
⇒ ∠COD = 55°
3. In Fig., AB is a chord of the circle and AOC is its diameter such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT is equal to
(A) 65°
(B) 60°
(C) 50°
(D) 40°
Solution
As given in the question,
A circle with centre O, diameter AC and
∠ACB = 50°
AT is a tangent to the circle at point A
As, angle in a semicircle is a right angle
∠CBA = 90°
So using angle sum property of a triangle,
∠ACB + ∠CAB + ∠CBA = 180°
⇒ 50° + ∠CAB + 90° = 180°
⇒ ∠CAB = 40° ...(1)
As, tangent to at any point on the circle is perpendicular to the radius through point of contact,
We get,
OA ⏊ AT
⇒ ∠OAT = 90°
⇒ ∠OAT + ∠BAT = 90°
⇒ ∠CAT + ∠BAT = 90°
⇒ 40° + ∠BAT = 90° [from equation (1)]
⇒ ∠BAT = 50°
4. From a point P which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is
(A) 60 cm^{2}
(B) 65 cm^{2}
(C) 30 cm^{2}
(D) 32.5 cm^{2}
(B) 65 cm^{2}
(C) 30 cm^{2}
(D) 32.5 cm^{2}
Solution
Construction: Draw a circle of radius 5 cm with center O.
Let P be a point at a distance of 13 cm from O.
Draw a pair of tangents, PQ and PR.
OQ = OR = radius = 5cm ...equation (1)
And OP = 13 cm
Also, tangent to at any point on the circle is perpendicular to the radius through point of contact,
We get,
OQ ⏊ PQ and OR ⏊PR
△POQ and △POR are rightangled triangles.
By using Pythagoras Theorem in △PQO,
(Base)^{2} + (Perpendicular)^{2} = (Hypotenuse)^{2}
⇒ (PQ)^{2} + (OQ)^{2} = (OP)^{2}
⇒ (PQ)^{2} + (5)^{2} = (13)^{2}
⇒ (PQ)^{2} + 25 = 169
⇒ (PQ)^{2} = 144
⇒ PQ = 12 cm
⇒ (PQ)^{2} = 144
⇒ PQ = 12 cm
Tangents through an external point to a circle are equal.
So,
PQ = PR = 12 cm ...(2)
Therefore,
Area of quadrilateral PQRS, A = area of △POQ + area of △POR
Area of right angled triangle = 1⁄2 x base x height
A = (1/2 × OQ × PQ) + (1/2 × OR × PR)
⇒ A = (1/2 × 5 × 12) + (1/2 × 5 × 12)
⇒ A = 30 + 30
⇒ A = 60 cm^{2}
5. At one end A of a 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 adistance 8 cm from A is
(A) 4 cm
(B) 5 cm
(C) 6 cm
(D) 8 cm
As given the question,
Radius of circle,
AO = OC = 5cm
AM = 8CM
⇒ AM = OM + AO
⇒ OM = AM  AO
Putting these values in the equation,
OM = (85)
= 3CM
OM is perpendicular to the chord CD.
In ∆OCM < OMC=90°
By Pythagoras theorem,
OC^{2} = OM^{2} + MC^{2}
Therefore,
CD = 2 ×CM
= 8 cm
6. In Fig., 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
Solution
(C)
Join OA
We know that, the tangent at any point of a circle is perpendicular to the radius through the
cos 30° = OT/AT
⇒ √3/2 = AT/4
⇒ AT = 2√3 cm
⇒ √3/2 = AT/4
⇒ AT = 2√3 cm
7. In Fig., if O is the 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°
Solution
(A)
Given,
∠QPR = 50°
We know that, the tangent at any point of a circle is perpendicular to the radius through the
point of contact.
∠OPR = 90°
⇒ ∠OPQ + ∠QPR = 90°
⇒ ∠OPQ = 90° – 50°
= 40° [as, ∠QPR = 50°]
Now,
OP = OQ = radius of circle
∠OQP = ∠OPQ
= 40°
[Angles opposite to equal sides are equal]
In ∆OPQ,
∠O + ∠OPQ + ∠Q = 180° [Angle sum property]
⇒ ∠POQ = 180° – (40 + 40°)
= 180° – 80°
[ ∠OPQ = 40° = ∠Q]
⇒ ∠POQ = 100°
8. In Fig., if PA and PB are tangents to the circle with centre O such that ∠APB 50° , then ∠OAB is equal to
(A) 25°
(B) 30°
(C) 40°
(D) 50°
Solution
(A)
Given,
PA and PB are tangent lines.
PA = PB [as, Length of tangents drawn from an external point to a circle is equal]
Let,
∠PBA = ∠PAB = Î¸
In ∆PAB,
∠P + ∠A + ∠B = 180° [Angle sum property]
⇒ 50° + Î¸ + Î¸= 180°
⇒ 2 Î¸ = 180° – 50° = 130°
⇒ Î¸ = 65°
Also,
OA ⊥ PA
[as, tangent at any point of a circle is perpendicular to the radius through the point of contact]
So,
∠PAO = 90°
⇒ ∠PAB + ∠BAO = 90°
⇒ 65° + ∠BAO = 90°
⇒ ∠BAO = 90° – 65° = 25°
⇒ ∠OAB = 25°
9. If two tangents inclined at an angle 60° are drawn to a circle of radius 3 cm, then length of each tangent is equal to
(A) 3√3/2 cm
(B) 6 cm
(C) 3 cm
(D) 3√3 cm
Solution
(D)
Let P be an external point and a pair of tangents is drawn from point P such that the angle between two tangents is 60°.
Join OA and OP.
Also,
OP is a bisector line of ∠APC.
∠APO = ∠CPO = 30°
And,
OA ⊥ AP
[Tangent at any point of a circle is perpendicular to the radius through the point of contact.]
∠OAP = 90°
In right angled ∆OAP,
tan 30° = OA/AP
tan 30° = OA/AP
⇒ 1/√3 = 3/AP
⇒ AP = 3√3 cm
So, the length of each tangent is 3√3 cm.
10. In Fig., if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR = 70° , then ∠AQB is equal to
(A) 20°
(B) 40°
(C) 35°
(D) 45°
Solution
(B)
Given,
AB  PR
Therefore,
∠ABQ = ∠BQR = 70° [Alternate angles]
Also,
QD is perpendicular to AB and QD bisects AB.
In ∆QDA and ∆QDB,
∠QDA = ∠QDB [90° each]
AD = BD
QD = QD [Common side]
So,
∆ADQ ≅ ∆BDQ [by SAS congruency]
Therefore,
∠QAD = ∠QBD [CPCT] ...(i)
But,
∠QBD = ∠ABQ = 70°
⇒ ∠QAD = 70° [From (i)]
Now, in ∆ABQ,
∠A + ∠B + ∠AQB = 180° [Angle sum property]
⇒ ∠AQB = 180° – (70° + 70°) = 40°