Two given circles $x^2 + y^2 + ax + by + c = 0$ and $x^2 + y^2 + dx + ey + f = 0$ will intersect each other orthogonally,only when

The point $(3, -4)$ lies on both the circles $x^2 + y^2 - 2x + 8y + 13 = 0$ and $x^2 + y^2 - 4x + 6y + 11 = 0$. Then,the angle between the circles is

If $2 x^{2}+2 y^{2}+4 x+5 y+1=0$ and $3 x^{2}+3 y^{2}+6 x-7 y+3 k=0$ are orthogonal,then the value of $k$ is

The point at which the circles $x^2+y^2-4x-4y+7=0$ and $x^2+y^2-12x-10y+45=0$ touch each other is:

If the circles $x^{2}+y^{2}+2x+2ky+6=0$ and $x^{2}+y^{2}+2ky+k=0$ intersect orthogonally, then $k$ is equal to

If $P$ is the point of contact of the circles $x^2+y^2+4x+4y-10=0$ and $x^2+y^2-6x-6y+10=0$,and $Q$ is their external centre of similitude,then the equation of the circle with $P$ and $Q$ as the extremities of its diameter is