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 $A B$ . Prove that the equation of the chord is $h x + k y = (\frac{d^{2}}{2} - a^{2})$ .

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Solution

$P (h, k)$ lies on circle $∴ h^{2} + k^{2} = a^{2} . . . . . (1)$
$△ P A B$ is isosceles hence $A B$ is $⊥$ to $O P$ whose slope is $k / h$ $∴$ Slope of $A B$ is $- h / k$ and hence its equation can be taken as $h x + k y = λ . . . . . (2)$
If $p$ be perpendicular from $O (0, 0)$ to $A B$ , then
$p = \frac{λ}{\sqrt{(h^{2} + k^{2})}} = \frac{λ}{a} . . . . . (3)$
Now we have to find the value of $λ$ in terms of known quantities $a$ and $d$ . From rt. angled triangles $A M O$ and $A M P$ .
$A M^{2} = a^{2} - p^{2} = d^{2} - (a + p)^{2}$
or $2 a^{2} = d^{2} - 2 a p = d^{2} - 2 λ$ , by $(3)$
$∴ λ = \frac{d^{2}}{2} - a^{2}$ . Put in $(2)$ etc.

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