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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
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)]
∴ 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.
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