The equation of the circle whose radius is $3$ and which touches internally the circle $x^2+y^2-4x-6y-12=0$ at the point $(-1,-1)$ is

The equation of the circle passing through $(2,0)$ and $(0,4)$ and having the minimum radius 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

Suppose $Q$ is a point on the circle with centre $P$ and radius $1$,as shown in the figure. $R$ is a point outside the circle such that $QR = 1$ and $\angle QRP = 2^{\circ}$. Let $S$ be the point where the segment $RP$ intersects the given circle. Then,the measure of $\angle RQS$ equals $......^{\circ}$.

If the equation of the circle whose radius is $3$ units and which touches internally the circle $x^2+y^2-4x-6y-12=0$ at the point $(-1,-1)$ is $x^2+y^2+px+qy+r=0$,then $p+q-r=$

If $(1,3)$ is the midpoint of a chord of the circle $x^2+y^2-4x-8y+16=0$,then the area of the triangle formed by that chord with the coordinate axes is