Question

AB and AC are two chords of a circle of Radius 'r' such that AB=2AC of P and Q are the distances of AB and AC from the center. Then prove that 4q²=p²+3r²

Answer 1

Given, AB and AC are two chords of a circle of radius r

Again, AB = 2AC and the distances of AB and AC from the center are p and q.

Let AC = x, then AB = 2x

We know that the perpendicular drawn from the centre to the chord bisect the chord.

So, CL = x/2 and AM = x

Now, in ΔOLC,

r2 = q2 + (x/2)2

=> r2 = q2 + x2 /4

Again, in ΔOAM,

4q2 = 4(r2 - x2 /4) (using 1)

4r2 = 4r2 - x2

=> 4q2 = 4r2 - (r2 - p2 ) (using 2)

=> 4q2 = 4r2 - r2 + p2

=> 4q2 = 3r2 + p2

Answer 2

Hope this helps you thankyou for asking me the question