If the circle passing through $(3,5), (5,5)$ and $(3,-3)$ cuts the circle $x^2+y^2+2x+2fy=0$ orthogonally, then $f=$

If $d$ is the distance between the centres of two circles,$r_1$ and $r_2$ are their radii,and $d = r_1 + r_2$,then

The equations of three circles are $x^2 + y^2 - 12x - 16y + 64 = 0$,$3x^2 + 3y^2 - 36x + 81 = 0$,and $x^2 + y^2 - 16x + 81 = 0$. The coordinates of the point from which the lengths of the tangents drawn to each of the three circles are equal is:

If the circles $x^2+y^2-2 \lambda x-2 y-7=0$ and $3(x^2+y^2)-8 x+29 y=0$ are orthogonal,then $\lambda=$

The circumference of the circle $x^2 + y^2 - 2x + 8y - q = 0$ is bisected by the circle $x^2 + y^2 + 4x + 12y + p = 0$. Then $p + q$ is equal to

The number of common tangents that can be drawn to the circles $x^2+y^2-6x=0$ and $x^2+y^2+6x+2y+1=0$ is .....