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, prove that 4q² = p² + 3r².

Answer 1

Given: In a circle of radius r, there are two chords AB and AC such that AB = 2AC. Also, the distance of AB and AC from the centre are p and q, respectively.

To prove: 4q² = p² + 3r²

Proof: Let AC = a, then AB = 2a

From centre O, perpendicular is drawn to the chords AC and AB at M and N, respectively.

∴ `AM = MC = a/2`

AN = NB = a

In ΔOAM, AO² = AM² + MO²   ...[By Pythagoras theorem]

⇒ `AO^2 = (a/2)^2 + q^2`  ...(i)

In ΔOAN, use Pythagoras theorem,

AO² = (AN)² + (NO)²

⇒ AO² = (a)² + p²  ...(ii)

From equations (i) and (ii),

`(a/2)^2 + q^2 = a^2 + p^2`

⇒ `a^2/4 + q^2 = a^2 + p^2`

⇒ a² + 4q² = 4a² + 4p²   ...[Multiplying both sides by 4]

⇒ 4q² = 3a² + 4p²

⇒ 4q² = p² + 3(a² + p²)

⇒ 4q² = p² + 3r²   ...[In right angled ΔOAN, r² = a² + p²]

Hence proved.