The equation of the circle whose diameter is the line segment joining the points $(-4, 3)$ and $(12, -1)$ is:

If the equation of the circle passing through the point $(8,8)$ and having the lines $x+2y-2=0$ and $2x+3y-1=0$ as its diameters is $x^2+y^2+px+qy+r=0$,then $p^2+q^2+r=$

The lengths of the intercepts made by a circle $S$ on $X$ and $Y$-axes are $\frac{2 \sqrt{13}}{3}$ and $\frac{2 \sqrt{22}}{3}$ respectively. If the radius of the circle $S$ is $\frac{\sqrt{38}}{3}$ and its centre $C$ lies in the second quadrant,then $C=$

The centre of a circle is $(2, -3)$ and the circumference is $10 \pi$. Then its equation is

The centre and radius of the circle $x^2+y^2+8x+10y-8=0$ respectively are

Find the equation of a circle whose center lies on the $y$-axis,has a radius of $3$,and passes through the origin.