Question

Question 12
AB and AC are two chords of a circle of radius r such that AB = 2AC. If p an q are the distances of AB an AC from the centre, prove that $4q^2=p^2+3r^2$.

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

Proof
Let AC = a, then AB = 2a
From centre O, perpendicular is drawn to the chords AC and AB at M and N, respectively
$\therefore AM=MC=\frac{a}{2}$
AN = NB = a

In $\Delta$ OAM
$A{O}^2=A{M}^2+M{O}^2$ [by Pythagoras theorem]
$A{O}^2={\left(\frac{a}{2}\right)}^2+q^2$ .........(i)

In $\Delta$ OAN,
$A{O}^2=(AN{)}^2+(NO{)}^2$ [by Pythagoras theorem]
$A{O}^2=a^2+p^2$ .............(ii)

From Eqs. (i) and (ii)
${\left(\frac{a}{2}\right)}^2+q^2=a^2+p^2$
$\Rightarrow \frac{a^2}{4}+q^2=a^2+p^2$
$\Rightarrow a^2+4q^2=4a^2+4p^2$ [multiplying both sides by 4]
$\Rightarrow 4q^2=3a^2+4p^2$
$\Rightarrow 4q^2=p^2+3(a^2+p^2)$
$\Rightarrow 4q^2=p^2+3r^2$ [in right angled $\Delta OAN,r^2=a^2+p^2$]
Hence proved.