The equations of the tangents to the circle ${x^2} + {y^2} = 50$ at the points where the line $x + 7 = 0$ meets it, are
$7x \pm y + 50 = 0$
$7x \pm y - 5 = 0$
$y \pm 7x + 5 = 0$
$y \pm 7x - 5 = 0$
Let a circle $C$ touch the lines $L_{1}: 4 x-3 y+K_{1}$ $=0$ and $L _{2}: 4 x -3 y + K _{2}=0, K _{1}, K _{2} \in R$. If a line passing through the centre of the circle $C$ intersects $L _{1}$ at $(-1,2)$ and $L _{2}$ at $(3,-6)$, then the equation of the circle $C$ is
Suppose two perpendicular tangents can be drawn from the origin to the circle $x^2+y^2-6 x-2 p y+17=0$, for some real $p$. Then, $|p|$ is equal to
The tangent and the normal lines at the point $(\sqrt 3,1)$ to the circle $x^2 + y^2 = 4$ and the $x -$ axis form a triangle. The area of this triangle (in square units) is
Consider a circle $(x-\alpha)^2+(y-\beta)^2=50$, where $\alpha, \beta>0$. If the circle touches the line $y+x=0$ at the point $P$, whose distance from the origin is $4 \sqrt{2}$ , then $(\alpha+\beta)^2$ is equal to................
If the ratio of the lengths of tangents drawn from the point $(f,g)$ to the given circle ${x^2} + {y^2} = 6$ and ${x^2} + {y^2} + 3x + 3y = 0$ be $2 : 1$, then