If the equation of the circle of radius $3$ units which touches the circle $x^2+y^2+6x-8y-11=0$ externally at $(3,0)$ is $x^2+y^2+2gx+2fy+c=0$,then $3g-4f+c=$

The lengths of tangents from a fixed point to three circles of a coaxial system are $t_1, t_2, t_3$. If $P, Q,$ and $R$ are the centers of these circles,then $QRt_1^2 + RPt_2^2 + PQt_3^2$ is equal to

$(a, 0)$ and $(b, 0)$ are centers of two circles belonging to a coaxial system of which the $y$-axis is the radical axis. If the radius of one of the circles is $r$,then the radius of the other circle is:

The equation of a circle passing through the origin and co-axial to the circles $x^2 + y^2 = a^2$ and $x^2 + y^2 + 2ax = 2a^2$ is

If $(a, b)$ and $(c, d)$ are the internal and external centres of similitude of the circles $x^2+y^2+4x-5=0$ and $x^2+y^2-6y+8=0$ respectively,then $(a+d)(b+c)=$

The centres of all circles passing through the points of intersection of the circles $x^2+y^2+2x-2y+1=0$ and $x^2+y^2-2x+2y-2=0$ and having radius $\sqrt{14}$ lie on the curve