If the equation of one tangent to the circle with center at $(2, -1)$ from the origin is $3x + y = 0$, then the equation of the other tangent through the origin is

The circle $x=5 \cos \theta, y=5 \sin \theta$ is bounded by the rectangle formed by the lines $x \pm 6=0$ and $y \pm 6=0$. The area of the triangle that lies inside the rectangle,which is formed by the tangent at $P\left(\frac{2 \pi}{3}\right)$ to the circle with two of the above given lines,is

The equation of the normal at $(1, 1)$ to the circle $x^2 + y^2 - x - 3y - 4 = 0$ is

The tangent to the circle $x^2+y^2=5$ at $(1,-2)$ also touches the circle $x^2+y^2-8x+6y+20=0$. The coordinates of the corresponding point of contact are:

Let a circle $C$ of radius $5$ lie below the $x$-axis. The line $L_{1}: 4x + 3y + 2 = 0$ passes through the centre $P$ of the circle $C$ and intersects the line $L_{2}: 3x - 4y - 11 = 0$ at $Q$. The line $L_{2}$ touches $C$ at the point $Q$. Then the distance of $P$ from the line $5x - 12y + 51 = 0$ is

The line $x \cos \alpha + y \sin \alpha = p$ will be a tangent to the circle $x^2 + y^2 - 2ax \cos \alpha - 2ay \sin \alpha = 0$,if $p = $