If the circles of same radius a and centers at $(2, 3)$ and $(5, 6)$ cut orthogonally, then $a =$
$1$
$2$
$3$
$4$
Coordinates of the centre of the circle which bisects the circumferences of the circles
$x^2 + y^2 = 1 ; x^2 + y^2 + 2x - 3 = 0$ and $x^2 + y^2 + 2y - 3 = 0$ is
The equation of the circle through the point of intersection of the circles ${x^2} + {y^2} - 8x - 2y + 7 = 0$, ${x^2} + {y^2} - 4x + 10y + 8 = 0$ and $(3, -3)$ is
If two circles ${(x - 1)^2} + {(y - 3)^2} = {r^2}$ and ${x^2} + {y^2} - 8x + 2y + 8 = 0$ intersect in two distinct points, then
The line $L$ passes through the points of intersection of the circles ${x^2} + {y^2} = 25$ and ${x^2} + {y^2} - 8x + 7 = 0$. The length of perpendicular from centre of second circle onto the line $L$, is
The circles ${x^2} + {y^2} + 4x + 6y + 3 = 0$ and $2({x^2} + {y^2}) + 6x + 4y + C = 0$ will cut orthogonally, if $C$ equals