Question

In a circle with centre P, chord AB is parallel to a tangent and intersects the radius drawn from the point of contact to its midpoint. If AB = `16sqrt(3)`, then find the radius of the circle.

Answer 1

Given: Chord AB || tangent XY

AB = `16sqrt(3)` units

PQ is radius of the circle.

PC = CQ

To find: Radius of the circle, i.e., l(PQ)

Construction: Draw seg PB.

In given figure, ∠PQY = 90°   ...(i) [Tangent theorem]

Chord AB || line XY   ...[Given]

∴ ∠PCB ≅ ∠PQY   ...[Corresponding angles]

∴ ∠PCB = 90°   ...(ii) [From (i)]

Now `CB = 1/2 AB`

∴ `CB = 1/2 xx 16sqrt(3)`   ...`[("A perpendicular drawn from the"),("centre of a circle on its chord"),("bisects the chord")]`

CB = `8sqrt(3)` units   ...(iii)

Let the radius of the circle be x units   ...(iv)

∴ PQ = x

∴ `PC = 1/2  PQ`   ...[PC = CQ, P–C–Q]

∴ `PC = 1/2 x`   ...(v)

In ∆PCB,

∠PCB = 90°   ...[From (ii)]

∴ PB² = PC² + CB²    ...[Pythagoras theorem]

∴ `x^2 = (1/2 x)^2 + (8sqrt(3))^2`   ...[From (iii), (iv) and (v)]

∴ `x^2 = x^2/4 + 64 xx 3`

∴ 4x² = x² + 192 × 4

∴ 4x² – x² = 768

∴ 3x² = 768

∴ `x^2 = 768/3`

∴ x² = 256

∴ x = 16 units   ...[Taking square root of both sides]

∴ The radius of the circle is 16 units.