If the tangent to the circle ${x^2} + {y^2} = {r^2}$ at the point $(a, b)$ meets the coordinate axes at the point $A$ and $B$, and $O$ is the origin, then the area of the triangle $OAB$ is

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 $x^2 + y^2 -6x+2y = 0$.

The equation of three circles are ${x^2} + {y^2} – 12x – 16y + 64 = 0,$ $3{x^2} + 3{y^2} – 36x + 81 = 0$ and ${x^2} + {y^2} – 16x + 81 = 0.$ The co-ordinates of the point from which the length of tangent drawn to each of the three circle is equal is

Let the tangents at two points $A$ and $B$ on the circle $x ^{2}+ y ^{2}-4 x +3=0$ meet at origin $O (0,0)$. Then the area of the triangle of $OAB$ is.

If the equation of the tangent to the circle ${x^2} + {y^2} – 2x + 6y – 6 = 0$ parallel to $3x – 4y + 7 = 0$ is $3x – 4y + k = 0$, then the values of $k$ are