The line $x-2=0$ cuts the circle $x^2+y^2-8x-2y+8=0$ at $A$ and $B$. The equation of the circle passing through the points $A$ and $B$ and having the least radius is

The line $Ax + By + C = 0$ cuts the circle $x^2 + y^2 + ax + by + c = 0$ at points $P$ and $Q$,and the line $A'x + B'y + C' = 0$ cuts the circle $x^2 + y^2 + a'x + b'y + c' = 0$ at points $R$ and $S$. If the four points $P, Q, R,$ and $S$ are concyclic,then $D = \left| {\begin{array}{*{20}{c}}{a - a'}&{b - b'}&{c - c'}\\A&B&C\\{A'}&{B'}&{C'}\end{array}} \right| = $

If $y = 2x$ is a chord of the circle $x^2 + y^2 - 10x = 0$,then the equation of the circle of which this chord is a diameter is:

Find the equation of a circle with radius $5$ units and touching the circle $x^2+y^2-2x-4y-20=0$ at the point $(5,5)$.

The circle passing through the intersection of the circles $x^{2}+y^{2}-6x=0$ and $x^{2}+y^{2}-4y=0$,having its centre on the line $2x-3y+12=0$,also passes through the point:

If the radical centre of the three circles $x^2+y^2=1$,$x^2+y^2-2x-3=0$,and $x^2+y^2-2y-3=0$ is $C(\alpha, \beta)$ and $r$ is the sum of the radii of the given circles,then the equation of the circle with $C(\alpha, \beta)$ as centre and $r$ as radius is: