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.
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.
Hint
Assume that $A(x_A,y_A),~~B(x_B,y_B)$ be the points of intersection the lin with the circle. So we have: $$(x_A-m)^2+(y_A-n)^2=d^2=(x_B-m)^2+(y_B-n)^2$$ From this identity, conclude that the slope of the line which is $\frac{y_A-y_B}{x_A-x_B}$ is $-m/n$. Now use the equality $(x_A-m)^2+(y_A-n)^2=d^2$ for the second one to find the equation of the line.
Let $O(0,0)$ be the origin and center of the circle. Let $A$ be the point $(m,n)$ and $P(x,y)$ be a point on the line you want.
Since the line is going to be perpendicular to $(OA)$, the equation has to look like $OA.OP = f(d)$, and we need to find out what $f(d)$ is.
Now, suppose $P$ is one of the intersection of that line with the circle, so that the distance $AP$ is $d$, and the distance $OP$ is $a$.
Then $d^2 = AP.AP = (AO+OP).(AO+OP) = 2a^2 + 2AO.OP$, hence $OA.OP = a^2 - \frac {d^2}2$.
So we must have $f(d) = a^2 - \frac{d^2}2$ and we have determined the equation of the line.
Where does the centre of the circle go in your diagram?
Since it is also equidistant from the two points, it lies on the same perpendicular.
So the slope of the perpendicular is $n/m$.
