The line $(x - a)\cos \alpha + (y - b)\sin \alpha = r$ will be a tangent to the circle $(x - a)^2 + (y - b)^2 = r^2$:

The equation of the normal to the circle $x^2+y^2+6x+4y-3=0$ at the point $(1, -2)$ is:

What is the equation of the tangent to the curve $x^2(x - y) + a^2(x + y) = 0$ at the origin?

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

If the line $3x - 4y - k = 0 (k > 0)$ touches the circle $x^2 + y^2 - 4x - 8y - 5 = 0$ at $(a, b)$,then $k + a + b$ is equal to:

$3x + 4y - 43 = 0$ is a tangent to the circle $S \equiv x^2 + y^2 - 6x + 8y + k = 0$ at a point $P$. If $C$ is the centre of the circle and $Q$ is a point which divides $CP$ in the ratio $-1:2$,then the power of the point $Q$ with respect to the circle $S = 0$ is