If $\left(0, \frac{3}{4}\right)$ is the radical centre of the circles $S \equiv x^2+y^2+\alpha x+6 y=0, S \equiv x^2+y^2+2 \alpha x+\alpha y+6=0$ and $S^{\prime \prime} \equiv x^2+y^2+6 \alpha x-\alpha y+3=0$, then the distance between the radical centre and the centre of the circle $S^{\prime}=0$ is

Let the slope of a diameter $A C$ of a circle of radius 25 units be $\frac{3}{4}$. If $(3,2)$ is the centre of the circle, $A=\left(x_1, y_1\right)$ and $C=\left(x_2, y_2\right)$, then $\frac{x_1 x_2}{y_1 y_2}=$

A circle passes through the points $(1,2)$, $(3,4)$. If its centre lines on the line $x-y+3=0$, then its radius is equal to

A line drawn through the point $A(5,7)$ cut the circle $x^2+y^2-36=0$ at the points $P$ and $Q$. Then, $A P \cdot A Q=$