Question

Find the co-ordinates of the point from which tangents are drawn to the circle $x^2+y^2-6x-4y+3=0$ such that mid-point of its chord of contact is $(1,1)$.

Answer 1

If the required point be $(\alpha ,\beta )$, then equation of $C.C.$ is
$\alpha x+\beta y-3(x+\alpha )-2(y+\beta )+3=0$
or $x(\alpha -3)+y(\beta -2)+(-3\alpha -2\beta +3)=0.....\left(1\right)$
Since $(1,1)$ is its mid-point then $T=S_1$.
Its equation is
$1.x+1.y-3(x+1)-2(y+1)+3=S_1=-5$
or $2x+y-3=0.....\left(2\right)$
Comparing $\left(1\right)$ and $\left(2\right)$,
$\frac{\alpha -3}{2}=\frac{\beta -2}{1}=\frac{3\alpha +2\beta -3}{3}$
$\therefore \alpha -2\beta =-1$ and $3\alpha -\beta =-3$
Solving, we get $\alpha =-1,\beta =0$.