The centre of the circle which intersects the circle $x^2+y^2-2x-2y-2=0$ orthogonally,passes through the point $(2,0)$,and touches the $X$-axis is:

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

If the circles $x^2+y^2-8x-8y+28=0$ and $x^2+y^2-8x-6y+25-\alpha^2=0$ have only one common tangent,then $\alpha=$

Let $A(1, 2)$ be the centre and $3$ be the radius of a circle $S$. Let $B(-1, -1)$ be the centre and $r$ be the radius of another circle $S^{\prime}$. If $\frac{\pi}{3}$ is the angle between the circles $S$ and $S^{\prime}$,then the number of possible values of $r$ is

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

If the equation of the circle which passes through the point $(1,1)$ and cuts both the circles $x^2+y^2-4x-6y+4=0$ and $x^2+y^2+6x-4y+15=0$ orthogonally is $x^2+y^2+2gx+2fy+c=0$,then $5g+2f+c=$