From a point P($\alpha$ ,$\beta$), two tangents PQ and PR are drawn to circle
x$^2$+y$^2$=a$^2$ Lets call this T
Find the equation of circumcircle of of triangle PQR. What I did was As follows Let the equation of circle be
x$^2$+y$^2$+2gx+2fy+c=0. Lets call this S
equation of line QR will be the radical axis of those two circles. Which is
x$\alpha$+y$\beta$-a$^2$=0
So we can conclude T-S=x$\alpha$+y$\beta$-a$^2$=0 Equation of radical axis
Which produces multiple solutions(equation of radical axis could be multiplied to any constant).
Like:
x$^2$+y$^2$-$\alpha$x-$\beta$y=0.
x$^2$+y$^2$+$\alpha$x+$\beta$y-a$^2$=0
This is where i am stuck.
Note that
$$T = x^2+y^2 -a^2, \>\>\>\>\>S=\alpha x +\beta y -a^2$$
Then, the equation of circumcircle of of triangle PQR is $T-S=0$, i.e.
$$ x^2+y^2 -\alpha x -\beta y=0$$
where there is no ambiguity.