The equation of the circle of radius $5$ and touching the coordinate axes in the third quadrant is

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

The radius of the circle $r=12 \cos \theta+5 \sin \theta$ is

Circles are drawn through the point $(2,0)$ to cut intercepts of length $5$ units on the $X$-axis. If their centre lies in the first quadrant,then their equation is

Let the abscissae of the two points $P$ and $Q$ on a circle be the roots of $x^{2}-4x-6=0$ and the ordinates of $P$ and $Q$ be the roots of $y^{2}+2y-7=0$. If $PQ$ is a diameter of the circle $x^{2}+y^{2}+2ax+2by+c=0$,then the value of $(a+b-c)$ 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