TOPICS

Unit-10(Examples)

Circles

**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.

Given that : 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 . Prove that : AP = BP. Construction : Let us join OP. Proof : 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.

Given that : 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. Prove that : ∠ PTQ = 2 ∠ OPQ. Proof : Let ∠ PTQ = θ Now, by Theorem 10.2, TP = TQ. So, TPQ is an isosceles triangle. Therefore, ∠ TPQ = ∠ TQP = 1/2 (180° - θ) = 90° - θ/2 Also, by Theorem 10.1, ∠ OPT = 90° So, ∠ OPQ = ∠ OPT – ∠ TPQ = 90° - (90° - θ/2) = θ/2 = ∠ PTQ/2 Hence, ∠ OPQ = ∠ PTQ/2 Therefore, ∠ 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. Find the length TP.

Join OT. Let it intersect PQ at the point R. Then Δ TPQ is isosceles and TO is the angle bisector of ∠ PTQ. So, OT ⊥ PQ and therefore, OT bisects PQ which gives PR = RQ = 4 cm. Also, OR = √OP² - PR² = √5² - 4² = 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 TP/5 = 4/3 TP = 20/3 cm

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