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 & AC from the centre are p & q, respectively.

To prove : $\large\sf { \:\:\:\:\:\:\:\:\:\:\:\:\: 4q^{2} = p^{2} + 3r^{2}}$

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

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

$\large\sf { \therefore \:\:\:\:\:\:\:\:\:\:\:\:\: AM = MC = \frac{a}{2}}$

$\large\sf { \:\:\:\:\:\:\:\:\:\:\:\:\: AN = NB = a}$

$\large\sf { In \Delta OAM \:\:\:\:\:\:\:\:\:\: AO^{2} = AM^{2} + MO^{2}}$ [ by Pythagoras theorem]

$\large\sf { \implies \:\:\:\:\:\:\:\:\:\:\:\:\: AO^{2} = \Big\lgroup\sf \frac{a}{2} \Big\rgroup ^{2} + q^{2}}$ ...(1)

$\large\sf { In \Delta OAN,}$ use Pythagoras theorem,

$\large\sf { \:\:\:\:\:\:\:\:\:\:\:\:\: AO^{2} = (a)^{2} + p^{2}}$ .....(2)

from eqns(1)&(2) ,

$\large\sf{ \:\:\:\:\:\:\:\:\:\:\:\:\: \Big\lgroup \frac{a}{2} \Big\rgroup + q^{2} = a^{2} + p^{2}}$

$\large\sf { \:\:\:\:\:\:\:\:\:\:\:\:\: \frac{a^{2}}{4} + q^{2} = a^{2} + p^{2}}$

$\large\sf { \:\:\:\:\:\:\:\:\:\:\:\:\: 4q^{2} = 3a^{2} + 4 p^{2}}$

$\large\sf { \:\:\:\:\:\:\:\:\:\:\:\:\: 4q^{2} = p^{2} + 3(a^{2} + p^{2})}$

$\large\sf{ \:\:\:\:\:\:\:\:\:\:\:\:\: \boxed{\implies 4q^{2} = p^{2} + 3r^{2}}}$

Answer 2

 Given :-

AB and AC are two chords of a circle of radius r .

such that AB = 2 AC , p and q are the distance of AB and AC from

 OM = p

ON = q

To prove : 4q² = p² +3r²

Construction : Join OA

Proof : OM and ON are the perpendicular distance of chords AB and AC from O.

∴ AM = 1/2 AB

AN = 1/2 AC

[ ∵ perpendicular from the centre bisects the chord ]

AM = 1/2 AB = 2AC /2 = AC = 2 AN

∴ AM = 2 AN......(1)

In right Δ AMO

AO² = AM² + OM²

r² = AM² + p²

AM² = r² - p²

( 2AN)² = r² - p².. from (1)

4AN ² = r² - p²

AN² = 1/4 (r² - p²)...(2)

In Δ ANO

AO² = AN² + ON²

r² = AN² + q²

AN² = r² - q²...(3)

From equation (2) and (3), we get

1/4 (r²⁻p²) = (r²- q²)

r² ⁻p² = 4 ( r² - q ² )

4q² = 4r² -r² + p²

4q² = 3r² + p²

Hence proved