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
$a + b + c = d + e + f$
$ad + be = c + f$
$ad + be = 2c + 2f$
$2ad + 2be = c + f$
If the circles ${x^2}\, + {y^2}\, - 16x\, - 20y\, + \,164\,\, = \,\,{r^2}$ and ${(x - 4)^2} + {(y - 7)^2} = 36$ intersect at two distinct points, then
The number of common tangents to the circles ${x^2} + {y^2} - 4x - 6y - 12 = 0$ and ${x^2} + {y^2} + 6x + 18y + 26 = 0$ 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
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 co-axial system of circles given by ${x^2} + {y^2} + 2gx + c = 0$ for $c < 0$ represents