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

Answer 1

Let $AC=a$ units. Therefore $AB=2a$ units

Let $OM\perp AC$ and $ON\perp AB$.

Therefore,

$AN=BN=a$ units and $AM=MC=\frac{a}{2}$units

Now, consider right $\Delta ANO$. By Pythagoras theorem,

$\Rightarrow A{O}^2=A{N}^2+O{N}^2$

$\Rightarrow r^2+a^2+p^2$

$\Rightarrow a^2=r^2-p^2$...(i)

Now, consider right $\Delta AMO$. By Pythagoras theorem,

$\Rightarrow A{O}^2=A{M}^2+O{M}^2$

$\Rightarrow r^2=\frac{a^2}{4}+q^2$

From (i), we get,

$\Rightarrow r^2=\frac{r^2-p^2}{4}+q^2$

$\Rightarrow 4r^2=r^2-p^2+4q^2$

$\Rightarrow 4q^2=3r^2+p^2$

Hence, proved.