Let $x^{2}+y^{2}+Ax+By+C=0$ be a circle passing through $(0,6)$ and touching the parabola $y=x^{2}$ at $(2,4)$. Then $A+C$ is equal to

Prove that the curves $y^{2}=4x$ and $x^{2}+y^{2}-6x+1=0$ touch each other at the point $(1,2)$.

If the distances from the origin to the centres of the three circles $x^2 + y^2 - 2\lambda_i x = c^2$ for $i = 1, 2, 3$ are in $G.P.$,then the lengths of the tangents drawn to them from any point on the circle $x^2 + y^2 = c^2$ are in

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 :

If the squares of the lengths of the tangents drawn from a point $P$ to the circles $x^{2} + y^{2} = a^2$,$x^2 + y^{2} = b^2$,and $x^{2} + y^{2} = c^{2}$ are in arithmetic progression,then:

$A$ pair of tangents are drawn from the origin to the circle $x^2 + y^2 + 20(x + y) + 20 = 0$. The equation of the pair of tangents is