Two orthogonal circles are such that area of one is twice the area of other. If radius of smaller circle is $r$, then distance between their centers will be -
$\sqrt 3 r$
$2r$
$\sqrt 5 r$
$3r$
The equation of the circle which passes through the point of intersection of circles ${x^2} + {y^2} - 8x - 2y + 7 = 0$ and ${x^2} + {y^2} - 4x + 10y + 8 = 0$ and having its centre on $y$ - axis, will be
If $P$ and $Q$ are the points of intersection of the circles ${x^2} + {y^2} + 3x + 7y + 2p - 5 = 0$ and ${x^2} + {y^2} + 2x + 2y - {p^2} = 0$ then there is a circle passing through $P, Q$ and $(1, 1)$ for:
The radical centre of the circles ${x^2} + {y^2} + 4x + 6y = 19,{x^2} + {y^2} = 9$ and ${x^2} + {y^2} - 2x - 2y = 5$ will be
The common tangent to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 + 6x + 8y - 24 = 0$ also passes through the point
If the two circles, $x^2 + y^2 + 2 g_1x + 2 f_1y = 0\, \& \,x^2 + y^2 + 2 g_2x + 2 f_2y = 0$ touch each then: