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

(N/A) Let there be a circle with centre $O$ and radius $r$. Let $AB$ and $AC$ be two chords such that $AB = 2 AC$.
Let $OL \perp AB$ and $OM \perp AC$. Given $OL = p$ and $OM = q$.
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
In right-angled $\Delta AOL$,by Pythagoras theorem:
$r^{2} = AL^{2} + p^{2} \Rightarrow AL^{2} = r^{2} - p^{2}$.
Since $L$ is the midpoint of $AB$,$AL = \frac{1}{2} AB$.
So,$(\frac{1}{2} AB)^{2} = r^{2} - p^{2} \Rightarrow \frac{1}{4} AB^{2} = r^{2} - p^{2} \Rightarrow AB^{2} = 4(r^{2} - p^{2})$.
Since $AB = 2 AC$,we have $(2 AC)^{2} = 4(r^{2} - p^{2}) \Rightarrow 4 AC^{2} = 4(r^{2} - p^{2}) \Rightarrow AC^{2} = r^{2} - p^{2} \quad \dots(1)$.
In right-angled $\Delta AOM$,by Pythagoras theorem:
$r^{2} = AM^{2} + q^{2} \Rightarrow AM^{2} = r^{2} - q^{2}$.
Since $M$ is the midpoint of $AC$,$AM = \frac{1}{2} AC$.
So,$(\frac{1}{2} AC)^{2} = r^{2} - q^{2} \Rightarrow \frac{1}{4} AC^{2} = r^{2} - q^{2} \Rightarrow AC^{2} = 4(r^{2} - q^{2}) \quad \dots(2)$.
Equating $(1)$ and $(2)$:
$r^{2} - p^{2} = 4(r^{2} - q^{2})$
$r^{2} - p^{2} = 4r^{2} - 4q^{2}$
$4q^{2} = 4r^{2} - r^{2} + p^{2}$
$4q^{2} = 3r^{2} + p^{2}$.
Hence,$4q^{2} = p^{2} + 3r^{2}$ is proved.