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 number of common tangents of the circles given by $x^2 +y^2 - 8x - 2y + 1 = 0$ and $x^2 + y^2 + 6x + 8y = 0$ is
Two circles ${S_1} = {x^2} + {y^2} + 2{g_1}x + 2{f_1}y + {c_1} = 0$ and ${S_2} = {x^2} + {y^2} + 2{g_2}x + 2{f_2}y + {c_2} = 0$ cut each other orthogonally, then
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
Radical axis of the circles $3{x^2} + 3{y^2} - 7x + 8y + 11 = 0$ and ${x^2} + {y^2} - 3x - 4y + 5 = 0$ is
The two circles ${x^2} + {y^2} - 2x + 22y + 5 = 0$ and ${x^2} + {y^2} + 14x + 6y + k = 0$ intersect orthogonally provided $k$ is equal to