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
$4$
$1$
$2$
$3$
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 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
The minimum distance between any two points $P _{1}$ and $P _{2}$ while considering point $P _{1}$ on one circle and point $P _{2}$ on the other circle for the given circles' equations
$x^{2}+y^{2}-10 x-10 y+41=0$
$x^{2}+y^{2}-24 x-10 y+160=0$ is .........
Let the circles $C_1:(x-\alpha)^2+(y-\beta)^2=r_1^2$ and $C_2:(x-8)^2+\left(y-\frac{15}{2}\right)^2=r_2^2$ touch each other externally at the point $(6,6)$. If the point $(6,6)$ divides the line segment joining the centres of the circles $C_1$ and $C_2$ internally in the ratio $2: 1$, then $(\alpha+\beta)+4\left(r_1^2+r_2^2\right)$ equals
Let $C_1$ and $C_2$ be the centres of the circles $x^2 + y^2 -2x -2y -2 = 0$ and $x^2 + y^2 - 6x-6y + 14 = 0$ respectively. If $P$ and $Q$ are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral $PC_1QC_2$ is ............. $\mathrm{sq. \, units}$