Tangent to the circle $x^2 + y^2$ = $5$ at the point $(1, -2)$ also touches the circle $x^2 + y^2 -8x + 6y + 20$ = $0$ . Then its point of contact is 

If a line, $y=m x+c$ is a tangent to the circle, $(x-3)^{2}+y^{2}=1$ and it is perpendicular to a line $\mathrm{L}_{1},$ where $\mathrm{L}_{1}$ is the tangent to the circle, $\mathrm{x}^{2}+\mathrm{y}^{2}=1$ at the point $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right),$ then

The two tangents to a circle from an external point are always

The tangent and the normal lines at the point $(\sqrt 3,1)$ to the circle $x^2 + y^2 = 4$ and the $x -$ axis form a triangle. The area of this triangle (in square units) is

If $\frac{x}{\alpha } + \frac{y}{\beta } = 1$ touches the circle ${x^2} + {y^2} = {a^2}$, then point $(1/\alpha ,\,1/\beta )$ lies on a/an