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]

∴ x² = `(1/2 x)^2 + (8sqrt(3))^2`   .....[From (iii), (iv) and (v)]

∴ x² = `x^2/4 + 64 xx 3`

∴ 4x² = `(x^2)/4 + 192`

∴ `(4x^2 – x^2)/4` = 192

∴ `(3x^2)/4` = 192

∴ x² = `192/3 xx 4`

∴ x² = 256

∴ `sqrt(x^2)` = `sqrt256`

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

∴ The radius of the circle is 16 units.