A circle passes through the points $(- 1, 1) , (0, 6)$ and $(5, 5)$ . The point$(s)$ on this circle, the tangent$(s)$ at which is/are parallel to the straight line joining the origin to its centre is/are :

Equation of the tangent to the circle, at the point $(1 , -1)$ whose centre is the point of intersection of the straight lines $x - y = 1$ and $2x + y= 3$ is

The normal at the point $(3, 4)$ on a circle cuts the circle at the point $(-1, -2)$. Then the equation of the circle is

The length of tangent from the point $(5, 1)$ to the circle ${x^2} + {y^2} + 6x - 4y - 3 = 0$, is

A circle $C_{1}$ passes through the origin $O$ and has diameter $4$ on the positive $x$-axis. The line $y =2 x$ gives a chord $OA$ of a circle $C _{1}$. Let $C _{2}$ be the circle with $OA$ as a diameter. If the tangent to $C _{2}$ at the point $A$ meets the $x$-axis at $P$ and $y$-axis at $Q$, then $QA : AP$ is equal to.

The line $y = mx + c$ will be a normal to the circle with radius $r$ and centre at $(a, b)$, if