A circle $S \equiv x^2+y^2-16=0$ intersects another circle $S^{\prime}=0$ of radius 5 units such that their common chord is of maximum length. If the slope of that chord is $\frac{3}{4}$, then the centre of such a circle $S^{\prime}=0$ is

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