AB and AC are two chords of a circle of radius r such that AB = 2AC if p - Maths - Circles

Question

AB and AC are two chords of a circle of radius r such that AB =
2AC if p and q are distances of AB and AC from the centre , prove
that 4q2 = p2 = 3r2

Ajanta Trivedi · Accepted Answer

given: AB and AC are two chords of a circle with center O. such that AB=2AC

p and q are the perpendicular distances of AB and AC from center O. i.e OM=p and ON=q

r is the radius of the circle.

TPT: $4 q^{2} = p^{2} + 3 r^{2}$

proof: join OA.

OM and ON are perpendicular distances of AB and AC from center O.

AN=AC/2 and AM=AB/2 [perpendicular from center to chord intersect at mid-point of the chord]

in right angled triangle ONA,

$O N^{2} + N A^{2} = O A^{2} q^{2} + N A^{2} = r^{2} N A^{2} = r^{2} - q^{2} ............ (1)$

in the right angled triangle OMA,

$O M^{2} + {A M}^{2} = O A^{2} p^{2} + {A M}^{2} = r^{2} {A M}^{2} = r^{2} - p^{2} ............ (2)$

since $A M = \frac{A B}{2} = \frac{2 A C}{2} = A C = 2 N A$

from (1) and (2)

$r^{2} - p^{2} = A M^{2} = (2 N A)^{2} = 4 N A^{2} r^{2} - p^{2} = 4 [r^{2} - q^{2}] r^{2} - p^{2} = 4 r^{2} - 4 q^{2} 4 q^{2} = 3 r^{2} + p^{2}$

which is the required result.

hope this helps you.

cheers!!!

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