Suppose that the circle $x^2+y^2+2gx+2fy+c=0$ has its centre on $2x+3y-7=0$ and cuts the circles $x^2+y^2-4x-6y+11=0$ and $x^2+y^2-10x-4y+21=0$ orthogonally. Then $5g-10f+3c=$

Let the set of all values of $r$, for which the circles $(x+1)^{2}+(y+4)^{2}=r^{2}$ and $x^{2}+y^{2}-4x-2y-4=0$ intersect at two distinct points, be the interval $(\alpha, \beta)$. Then $\alpha\beta$ is equal to

The point of intersection of the common tangents drawn to the circles $x^2+y^2-4x-2y+1=0$ and $x^2+y^2-6x-4y+4=0$ is:

The equation of the family of circles passing through the point of intersection of a circle $S = 0$ and a line $P = 0$ is given by:

If the lengths of the tangents drawn from a point $P$ to the circles $x^2+y^2-8x+40=0$,$5x^2+5y^2-25x+80=0$,and $x^2+y^2-8x+16y+160=0$ are equal,then the coordinates of point $P$ are:

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