The equation of the circle which passes through the intersection of ${x^2} + {y^2} + 13x - 3y = 0$and $2{x^2} + 2{y^2} + 4x - 7y - 25 = 0$ and whose centre lies on $13x + 30y = 0$ is
If the line $x\, cos \theta + y\, sin \theta = 2$ is the equation of a transverse common tangent to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 6 \sqrt{3} \,x - 6y + 20 = 0$, then the value of $\theta$ is :
$P, Q$ and $R$ are the centres and ${r_1},\,\,{r_2},\,\,{r_3}$ are the radii respectively of three co-axial circles, then $QRr_1^2 + RP\,r_2^2 + PQr_3^2$ is equal to
The point of contact of the given circles ${x^2} + {y^2} - 6x - 6y + 10 = 0$ and ${x^2} + {y^2} = 2$, is
Consider a circle $C_1: x^2+y^2-4 x-2 y=\alpha-5$.Let its mirror image in the line $y=2 x+1$ be another circle $C _2: 5 x ^2+5 y ^2-10 fx -10 gy +36=0$.Let $r$ be the radius of $C _2$. Then $\alpha+ r$ is equal to $......$.