The length of the shortest path that begins at the point $(2, 5)$,touches the $x-$axis,and then ends at a point on the circle $x^2 + y^2 + 12x - 20y + 120 = 0$.

The number of common tangents to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 6x - 8y = 24$ is

$A$ line is drawn through a fixed point $P(\alpha, \beta)$ to cut the circle $x^{2}+y^{2}=r^{2}$ at $A$ and $B$. Then $PA \cdot PB$ is equal to

For the four circles $M, N, O$ and $P$,the following four equations are given:
Circle $M: x^2 + y^2 = 1$
Circle $N: x^2 + y^2 - 2x = 0$
Circle $O: x^2 + y^2 - 2x - 2y + 1 = 0$
Circle $P: x^2 + y^2 - 2y = 0$
If the centre of circle $M$ is joined with the centre of circle $N$,the centre of circle $N$ is joined with the centre of circle $O$,the centre of circle $O$ is joined with the centre of circle $P$,and lastly,the centre of circle $P$ is joined with the centre of circle $M$,then these lines form the sides of a:

The circumference of a circle passing through the point $(4,6)$ with two normals represented by $2x - 3y + 4 = 0$ and $x + y - 3 = 0$ is (in $\pi$)

The line $4x + 3y - 4 = 0$ divides the circumference of a circle in the ratio $1:2$. If $C(5, 3)$ is the center of that circle,then the equation of the circle is