Let the tangent to the circle $C _{1}: x^{2}+y^{2}=2$ at the point $M (-1,1)$ intersect the circle $C _{2}$ : $( x -3)^{2}+(y-2)^{2}=5$, at two distinct points $A$ and $B$. If the tangents to $C _{2}$ at the points $A$ and $B$ intersect at $N$, then the area of the triangle $ANB$ is equal to

Tangents drawn from the point $P(1,8)$ to the circle $x^2+y^2-6 x-4 y-11=0$ touch the circle at the points $A$ and $B$. The equation of the circumcircle of the triangle $P A B$ is

The equations of the tangents drawn from the origin to the circle ${x^2} + {y^2} - 2rx - 2hy + {h^2} = 0$ are

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 

The line $x\cos \alpha + y\sin \alpha = p$will be a tangent to the circle ${x^2} + {y^2} - 2ax\cos \alpha - 2ay\sin \alpha = 0$, if $p = $

Statement $1$ : The only circle having radius $\sqrt {10} $ and a diameter along line $2x + y = 5$ is $x^2 + y^2 - 6x +2y = 0$.
Statement $2$ : $2x + y = 5$ is a normal to the circle $x^2 + y^2 -6x+2y = 0$.