Two orthogonal circles are such that the area of one is twice the area of the other. If the radius of the smaller circle is $r$,then the distance between their centers will be -

$(a, 0)$ and $(b, 0)$ are the centres 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

If the circle $S=0$ intersects the three circles $S_1 \equiv x^2+y^2+4x-7=0$,$S_2 \equiv x^2+y^2+y=0$ and $S_3 \equiv x^2+y^2+\frac{3}{2}x+\frac{5}{2}y-\frac{9}{2}=0$ orthogonally,then the radical axis of $S=0$ and $S_1=0$ is

$A$ circle $S = x^2 + y^2 + 2gx + 2fy + 4 = 0$ cuts the circle $x^2 + y^2 - 4x - 4y - 4 = 0$ orthogonally and makes an angle of $60^{\circ}$ with the circle $x^2 + y^2 + 4x + 4y + 4 = 0$. Then the radius of the circle $S = 0$ is

If $x-y+1=0$ meets the circle $x^2+y^2+y-1=0$ at $A$ and $B$,then the equation of the circle with $AB$ as diameter is

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