If the tangents drawn at the points $O(0,0)$ and $P(1+\sqrt{5}, 2)$ on the circle $x^{2}+y^{2}-2x-4y=0$ intersect at the point $Q$,then the area of the triangle $OPQ$ is equal to

Let the tangent to the circle $x^{2}+y^{2}=25$ at the point $R(3,4)$ meet the $x$-axis and $y$-axis at points $P$ and $Q$,respectively. If $r$ is the radius of the circle passing through the origin $O$ and having its centre at the incentre of the triangle $OPQ$,then $r^{2}$ 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

The equation of the normal to the circle $x^2+y^2-2x+4y-5=0$ at the point $(2,1)$ is:

The length of the tangent from a point $P(x_1, y_1)$ to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by:

$A$ circle with centre $(a, b)$ passes through the origin. The equation of the tangent to the circle at the origin is