Let $\theta$ be the angle between the circles $S \equiv x^2+y^2+2 x-2 y+c=0$ and $S^{\prime} \equiv x^2+y^2-6 x-8 y+9=0$. If $c$ is an integer and $\cos \theta=\frac{5}{16}$, then the radius of the circle $S=0$ is

If a circle $S$ passes through the origin and makes an intercept of length 4 units on the line $x=2$, then the equation of the curve on which the centre of $S$ lies is

A circle touches the line $2 x+y-10=0$ at $(3,4)$ and passes through the point $(1,-2)$. Then, a point that lies on the circle is

If $(a, b)$ is the common point for the circles $x^2+y^2-4 x+4 y-1=0$ and $x^2+y^2+2 x-4 y+1=0$, then $a^2+b^2=$