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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

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#### Solution

**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)]

∴ PB^{2} = PC^{2} + CB^{2} .....[Pythagoras theorem]

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

∴ x^{2} = `x^2/4 + 64 xx 3`

∴ 4x^{2} = `(x^2)/4 + 192`

∴ `(4x^2 – x^2)/4` = 192

∴ `(3x^2)/4` = 192

∴ x^{2} = `192/3 xx 4`

∴ x^{2 }= 256

∴ `sqrt(x^2)` = `sqrt256`

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

∴ The radius of the circle is 16 units.

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