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 circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$ bisects the circumference of the circle ${x^2} + {y^2} + 2g'x + 2f'y + c' = 0$, if
In the figure shown, radius of circle $C_1$ be $ r$ and that of $C_2$ be $\frac{r}{2}$ , where $r= \frac {1}{3} PQ,$ then length of $AB$ is (where $P$ and $Q$ being centres of $C_1$ $\&$ $C_2$ respectively)
The points of intersection of circles ${x^2} + {y^2} = 2ax$ and ${x^2} + {y^2} = 2by$ are
If a circle $C$ passing through $(4, 0)$ touches the circle $x^2 + y^2 + 4x - 6y - 12 = 0$ externally at a point $(1, -1),$ then the radius of the circle $C$ 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