Question

P is in the exterior of a circle at distance 34 from the centre 0. A line through P touches the circle at Q. PQ = 16, find the diameter of the circle.

Answer 1

We know, tangent of circle at point of contact making right angle with the radius.

Hence, using Pythagoras Theorem in ΔOQP

OQ^{2} +PQ^{2} =OP^{2}

OQ^{2} +16^{2} = 34^{2}

OQ^{2} =34^{2} -16^{2}

OQ^{2} = 30^{2}

Hence, OQ = 30

Here, OQ is radius so diameter is 2OQ = 60

Answer 2

Given P is the exterior of a Circle at

distance 34 from the centre O .

A line through P touches the Circle

at Q .

PQ = 16

From the rough diagram attached,

OP is perpendicular to PQ

[ radius , tangent relation ]

Now ,

In ∆OPQ , <Q = 90°

By Phythogarian theorem ,

OP² = PQ² + OQ²

=> 34² = 16² + OQ²

=> 34² - 16² = OQ²

=> ( 34 + 16 )( 34 - 16 ) = OQ²

=> 50 × 18 = OQ²

=> 900 = OQ²

=> OQ² = 30²

=> OQ = 30

Therefore ,

diameter ( d ) = 2 × radius

=> d = 2 × OQ

=> d = 2 × 30

=> d = 60

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