At which point does the line $x = 0$ touch the circle $x^2 + y^2 - 2x - 6y + 9 = 0$?

Let the lines $(2-i)z = (2+i)\bar{z}$ and $(2+i)z + (i-2)\bar{z} - 4i = 0$ (where $i^2 = -1$) be normal to a circle $C$. If the line $iz + \bar{z} + 1 + i = 0$ is tangent to this circle $C$,then its radius is

The line $y = x + c$ will intersect the circle $x^2 + y^2 = 1$ in two coincident points,if

The point of contact of the tangent to the circle $x^2 + y^2 = 5$ at the point $(1, -2)$ which also touches the circle $x^2 + y^2 - 8x + 6y + 20 = 0$ is:

The equation of the normal to the circle $x^2 + y^2 = 9$ at the point $\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$ is

$A$ circle $C_{1}$ passes through the origin $O$ and has a diameter of $4$ on the positive $x$-axis. The line $y = 2x$ intersects the circle $C_{1}$ at $O$ and $A$. 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 the $y$-axis at $Q$,then the ratio $QA : AP$ is equal to: