The point of contact of the given circles $x^2 + y^2 - 6x - 6y + 10 = 0$ and $x^2 + y^2 = 2$ is

The straight line $x \cos \alpha + y \sin \alpha = p$ cuts the circle $x^2 + y^2 - a^2 = 0$ at $A$ and $B$. Then the equation of the circle having $AB$ as diameter is

The angle of intersection of the circles $x^2 + y^2 - x + y - 8 = 0$ and $x^2 + y^2 + 2x + 2y - 11 = 0$ is

Let $a=1+i$ and $z=x+iy$. If the curve $z\bar{z}+az+\bar{a}\bar{z}-4=0$ is cut by the straight line $(z+\bar{z})-i(z-\bar{z})+2=0$ at two points $A$ and $B$, then the equation of the circle passing through the origin, $A$ and $B$ is

If the circles $x^2+y^2-4x+2fy+1=0$ and $x^2+y^2+2gx-4y-1=0$ cut orthogonally,then $r_1^2+r_2^2-8=$

The circles $x^2 + y^2 + 2gx + 2fy = 0$ and $x^2 + y^2 + 2g'x + 2f'y = 0$ touch each other externally if: