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

Given AB and AC are two chords of a circle of radius r such that AB = 2AC.

P and q are perpendicular distances of AB and AC from centre O.

⇒ OM = p and ON = q

We have to prove that 4q² = p² + 3r²

Proof:

We know that perpendicular from centre to chord intersect at mid-point of the chord.

Consider ΔONA,

By Pythagoras Theorem,

⇒ ON² + NA² = OA²

⇒ q² + NA² = r²

⇒ NA² = r² – q² … (1)

Consider ΔOMA,

By Pythagoras Theorem,

⇒ OM² + AM² = OA²

⇒ p² + AM² = r²

⇒ AM² = r² – p² … (2)

⇒ AM = = = AC = 2NA

From (1) and (2),

⇒ r² – p² = AM² = (2NA)² = 4NA²

⇒ r² – p² = 4 (r² – q²)

⇒ r² – p² = 4r² – 4q²

∴ 4q² = 3r² + p²

Hence proved.