NCERT Solutions Class 10th Maths Chapter – 10 Circle Examples

Example 1. Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.

Solution: We are given two concentric circles C₁ and C₂ with centre O and a chord AB of the larger circle C₁ which touches the smaller circle C₂ at the point P. We need to prove that AP BP.
Let us join OP. Then, AB is a tangent to C₂ at P and OP is its radius. Therefore, by Theorem 10.1, OP ⊥ AB
Now AB is a chord of the circle C₁ and OP ⊥ AB. Therefore, OP is the bisector of the chord AB, as the perpendicular from the centre bisects the chord,
i.e., AP = BP

Example 2. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2 ∠OPQ.

Solution: We are given a circle with centre O, an external point T and two tangents TP and TQ to the circle, where P, Q are the points of contact We need to prove that
∠PTQ = 2 ∠OPQ
∠PTQ = 0
Now, by Theorem 10.2, TP = TQ. So, TPQ is an isosceles triangle.
Therefore,
∠TPQ = ∠TQP = 1/2 (180° – 0) = 90° -1/2 0
Also, by Theorem 10.1, ∠OPT = 90°
So,
∠OPQ = ∠OPT – ∠TPQ = 90° -[90° -1/2 0]
1/2 0 = 1/2 ∠PTQ
This gives ∠PTQ = 2 ∠OPQ

Example 3. PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T (see Fig. 10.10). Find the length TP.

Solution: Join OT. Let it intersect PQ at the point R. Then A TPQ is isosceles and TO is the angle bisector of PTQ. So, OT 1 PQ and therefore, OT bisects PQ which gives PR=RQ = 4 cm. Also, OR = √OP2 – PR² = √5² – 42 cm = 3 cm.
Now, ∠TPR + ∠RPO = 90° = ∠TPR + ∠PTR (Why?)
So, ∠RPO = ∠PTR
Therefore, right triangle TRP is similar to the right triangle PRO by AA similarity.
This gives TP/PO = RP/RO, i.e., TP/5 = 4/3 = or TP = 20/3cm.
Note: TP can also be found by using the Pythagoras Theorem, as follows:
Let
TP = x and TR = y. Then
x² = y² +16 (Taking right A PRT) …1
x²+52 = (y+3)² (Taking right A OPT) …2
Subtracting (1) from (2), we get
25 = 6y – 7 or y = 32/6 = 16/3
Therefore, x2 = (16/3)2 + 16 = 16/9(16 + 9) = 16 × 25/9 [from (1)]
or x = 20/3