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Question

AB and AC are two chords of a circle of radius r such that AB=2AC. If p and q are the distances of AB and AC from the centre, then show that 4q2=p2+3r2.

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Solution

Consider the diagram shown below.

Given:

AB and AC are two chords of a circle with center O. Such that AB=2AC

p and q are distances of AB and AC from center O i.e., OM=p and ON=q

r is the radius of the circle

To prove that:

4q2=p2+3r2

Proof:

Join OA.

OM and ON are distances of AB and AC from center O.

Here,

AN=AC2 (perpendicular from center to chord intersect at mid-point of the chord)

AM=AB2 (perpendicular from center to chord intersect at mid-point of the chord)

In right angled ΔOMA,

OM2+AM2=OA2

p2+AM2=r2

AM2=r2p2 …… (1)

In right angled ΔONA,

ON2+AN2=OA2

q2+AN2=r2

AN2=r2q2 …… (2)

Since,

AM=AB2=2AC2=AC=2AN

From equations (1) and (2), we have

r2p2=AM2

r2p2=4AN2

r2p2=4[r2q2]

4q2=p2+3r2

LHS=RHS

Hence, proved.


1061351_426727_ans_c9e257420d10412c811e2a4ae75da9ea.PNG

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