Question

A point $P(h,k)$ of the circle $x^2+y^2=a^2$ is at a constant distance $d$ from the extremities of the chord $AB$. Prove that the equation of the chord is $hx+ky=(\frac{d^2}{2}-a^2)$.

Answer 1

$P(h,k)$ lies on circle $\therefore h^2+k^2=a^2.....\left(1\right)$
$△PAB$ is isosceles hence $AB$ is $\perp$ to $OP$ whose slope is $k/h$ $\therefore$ Slope of $AB$ is $-h/k$ and hence its equation can be taken as $hx+ky=\lambda .....\left(2\right)$
If $p$ be perpendicular from $O(0,0)$ to $AB$, then
$p=\frac{\lambda }{\sqrt{(h^2+k^2)}}=\frac{\lambda }{a}.....\left(3\right)$
Now we have to find the value of $\lambda$ in terms of known quantities $a$ and $d$. From rt. angled triangles $AMO$ and $AMP$.
$A{M}^2=a^2-p^2=d^2-(a+p{)}^2$
or $2a^2=d^2-2ap=d^2-2\lambda$, by $\left(3\right)$
$\therefore \lambda =\frac{d^2}{2}-a^2$. Put in $\left(2\right)$ etc.