Chapter 9 Circles NCERT Exemplar Solutions Exercise 9.3 Class 10 Maths
Chapter Name  NCERT Maths Exemplar Solutions for Chapter 9 Circles Exercise 9.3 
Book Name  NCERT Exemplar for Class 10 Maths 
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Related Study  NCERT Solutions for Class 10 Maths 
Exercise 9.3 Solutions
Short Answer Questions
1. Out of the two concentric circles, the radius of the outer circle is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle.
Solution
Let C_{1} , and C_{2} be the two circles having same centre O. AC is a chord which touches C_{1} at point D.
Join OD.
Also, OD ⊥ AC
AD = DC = 4 cm
[Perpendicular line OD bisects the chord]
In right angled ∆AOD,
OA^{2} = AD^{2} + OD^{2} [By Pythagoras theorem]
⇒ OD^{2} = 5^{2} – 4^{2}⇒ OD^{2} = 25 – 16 = 9
⇒ OD = 3 cm
Radius of the inner circle is OD = 3 cm
2. Two tangents PQ and PR are drawn from an external point to a circle with centre O. Prove that QORP is a cyclic quadrilateral.
Solution
We know that,
Radius ⊥ Tangent = OR ⊥ PR
∠ORP = 90°
Similarly,
Radius ⊥ Tangent = OQ ⊥PQ
∠OQP = 90°
In quadrilateral ORPQ,
Sum of all interior angles = 360°
∠ORP + ∠RPQ+ ∠PQO + ∠QOR = 360°
⇒ 90° + ∠RPQ + 90° + ∠QOR = 360°
So,
∠O + ∠P = 180°
PROQ is a cyclic quadrilateral.
3. If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that ∠DBC = 120°, prove that BC + BD = BO, i.e., BO = 2BC.
Solution
As given in the question,
By RHS rule,
Î”OBC and Î”OBD are congruent {By CPCT}
∠OBC and ∠ OBD are equal
Therefore,
∠OBC = ∠OBD =60°
In triangle OBC,
cos 60°= BC/OB
⇒ 1⁄2 = BC/OB
⇒ OB = 2BC
Hence proved.
4. Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.
Solution
Let us take the lines be l_{1} and l_{2} .
Assume that O touches l1 and l2 at M and N,
So,
OM = ON (Radius of the circle)
Therefore,
From the centre ”O” of the circle, it has equal distance from l1 & l2.
In Î” OPM & OPN,
OM = ON (Radius of the circle)
∠OMP = ∠ONP (As, Radius is perpendicular to its tangent)
OP = OP (Common sides)
Therefore,
Î” OPM = Î”OPN (SSS congruence rule)
By C.P.C.T,
∠MPO = ∠NPO
So, l bisects ∠MPN.
Hence, O lies on the bisector of the angle between l1 & l2 .
Also,
Centre of a circle touching two intersecting lines lies on the angle bisector of the lines.
5. In Fig. 9.13, AB and CD are common tangents to two circles of unequal radii. Prove that AB = CD.
Solution
As given in the question,
AB = CD
Construction: Produce AB and CD, to intersect at P.
Proof:
Consider the circle with greater radius.
Tangents drawn from an external point to a circle are equal
AP = CP ...(i)
Also,
Consider the circle with smaller radius.
Tangents drawn from an external point to a circle are equal
BP = BD ...(ii)
Subtract Equation (ii) from (i),
AP – BP = CP – BD
⇒ AB = CD
Hence Proved.
6. In Question 5 above, if radii of the two circles are equal, prove that AB = CD.
Solution
Join OO’
Since, OA = O’B [Given]
And,
∠OAB = ∠O’BA = 90°
[Tangent at any point of a circle is perpendicular to the radius at the point of contact]
Since, perpendicular distance between two straight lines at two different points is same.
AB is parallel to OO’
Also,
CD is parallel to OO’
AB  CD
Now,
∠OAB = ∠OCD = ∠O’BA = ∠O’DC = 90°
ABCD is a rectangle.
Hence,
AB = CD.