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 prove that 4q² = p² + 3r².

Answer 1

Consider O as the centre of the circle with radius r

So we get

OB = OC = r

Consider AC = x and AB = 2x

We know that OM ⊥ AB

So we get

OM = p

We know that ON ⊥ AC

So we get

ON = q

Consider △ OMB

Using the Pythagoras theorem

OB² = OM² + BM²

We know that the perpendicular from the centre of the circle bisects the chord

So we get

r² = p² + ((1/2) AB) ²

It can be written as

r² = p² + ¼ × 4x²

So we get

r² = p² + x² ……… (1)

Consider △ ONC

Using the Pythagoras theorem

OC² = ON² + CN²

We know that the perpendicular from the centre of the circle bisects the chord

So we get

r² = q² + ((1/2) AC) ²

It can be written as

r² = q² + x²/4

We get

q² = r² – x²/4

Multiplying the equation by 4

4q² = 4r² – x²

Substituting equation (1)

4q² = 4r² – (r² – p²)

So we get

4q² = 3r² + p²

Therefore, it is proved that 4q² = p² + 3r².