The equation of the circle having the lines ${x^2} + 2xy + 3x + 6y = 0$ as its normals and having size just sufficient to contain the circle $x(x - 4) + y(y - 3) = 0$is

Let the latus ractum of the parabola $y ^{2}=4 x$ be the common chord to the circles $C _{1}$ and $C _{2}$ each of them having radius $2 \sqrt{5}$. Then, the distance between the centres of the circles $C _{1}$ and $C _{2}$ is

The value of $'c'$ for which the set, $\{(x, y) | x^2 + y^2 + 2x \le 1 \} \cap \{(x, y) | x - y + c \ge 0\}$ contains only one point in common is :

A circle $C_1$ of radius $2$ touches both $x$ -axis and $y$ -axis. Another circle $C_2$ whose radius is greater than $2$ touches circle $C_1$ and both the axes. Then the radius of  circle $C_2$ is-

The two circles ${x^2} + {y^2} - 2x - 3 = 0$ and ${x^2} + {y^2} - 4x - 6y - 8 = 0$ are such that

If the circles ${x^2} + {y^2} + 2ax + cy + a = 0$ and ${x^2} + {y^2} - 3ax + dy - 1 = 0$ intersect in two distinct points $P$ and $Q$ then the line $5x + by - a = 0$ passes through $P$ and $Q$ for