Q:

A proof in circles

The question is this:
Show that the equation of a straight line meeting the circle $ x^2 + y^2 = a^2 $ in two points at equal distances $d$ from a point $(m, n)$ on its circumference is 
$ mx + ny - a^2 + \dfrac {d^2} {2} $
I am confused. Hints please!
EDIT:
I tried to relate the perpendicular distance between the line and $(m, n)$ but then I seem to be getting nowhere.

A:

Let $p, q$ be the two intersections of the line with the circle:
$$x^2 + y^2 - a^2 = 0\tag{*1}$$
Since $p$ and $q$ are at distance $d$ from the point $(m,n)$, $p$, $q$ lies on another 
circle:
$$(x - m)^2 + (y - n)^2 - d^2 = 0\tag{*2}$$
$(*1) - (*2)$ implies $p, q$ also satisfy:
$$\begin{align}&2mx - m^2 + 2nx - n^2 - a^2 + d^2 = 0\\
\iff & 2mx + 2ny - 2 a^2 + d^2 = 0\\
\iff & mx + ny - a^2 + \frac{d^2}{2} = 0
\end{align}$$
This is the equation of a line and since two points determine a line, this is the equation we want.

