The equation of a circle passing through points of intersection of the circles ${x^2} + {y^2} + 13x – 3y = 0$ and $2{x^2} + 2{y^2} + 4x – 7y – 25 = 0$ and point $(1, 1)$ is

The value of $\lambda $, for which the circle ${x^2} + {y^2} + 2\lambda x + 6y + 1 = 0$, intersects the circle ${x^2} + {y^2} + 4x + 2y = 0$ orthogonally is

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)

Two given circles ${x^2} + {y^2} + ax + by + c = 0$ and ${x^2} + {y^2} + dx + ey + f = 0$ will intersect each other orthogonally, only when

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 $……$.