A circle and an equation

Geometry Level 3

$AB$ and $AC$ are two chords of a circle with centre $O$ and radius $r$ such that $AB=2AC$ . $p$ and $q$ are the distances of $AB$ and $AC$ respectively from the centre of the circle. Then which of the following is true?

4q^2= 3p^2 + r^2

3p^2= q^2+2r^2

3p^2= 2q^2 + 2r^2

4q^2= p^2 + 3r^2

1 solution

Yatin Khanna
Feb 28, 2016

- First of all, draw perpendiculars OM and ON on AB and AC respectively. - OM='p' - ON='q'

Now, by pythagoras' theorem:

OA^2 = OM^2 + MA^2 ======================> r^2 = p^2 + (1/2AB)^2 (perpendicular from the centre bisects a chord.)

Similarly in triangle ONA: r^2= q^2 + (1/2AC)^2
THIS IMPLIES THAT: p^2+ (1/2AB)^2= q^2 + (1/2AC)^2 ================> p^2 + AC^2 = q^2 + 1/4AC^2 (AB= 2AC, THUS 1/2 AB=AC) =========================> p^2+ 3/4AC^2 = q^2 =====================> 4p^2 + 3AC^2 = 4q^2 =====================> p^2 + 3p^2 + 3AC^2 = 4q^2 ===================> p^2 + 3(p^2 + AC^2) = 4q^2 ====================> p^2 + 3r^2= 4q^2 (In right angled triangle OMA p^2+ AC^2= OA^2 and OA = 'r'.)
THUS, 4q^2= p^2 + 3r^2

SORRY FOR THE SOLUTION BEING SO MESSY...ITS MY FIRST TIME AND I AM HAVING A BIT PROBLEM IN POSTING THE SOLUTION THE WAY I WANT..

A circle and an equation

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