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 the tangents at the points $A (4, -11)$ and $B (8, -5)$ on the circle $x^2 + y^2 - 3x + 10y - 15 = 0$ intersect at the point $C$. Then the radius of the circle,whose center is $C$ and the line joining $A$ and $B$ is its tangent,is equal to

$A$ pair of tangents are drawn to the circle $x^2+y^2+6x-4y-12=0$ from a point $P(-4,-5)$. The area enclosed between these tangents and the circle is:

If the straight line $4x + 3y + \lambda = 0$ touches the circle $2(x^2 + y^2) = 5$,then $\lambda$ is

Two tangents to the circle $x^{2}+y^{2}=4$ at the points $A$ and $B$ meet at $M(-4,0)$. The area of the quadrilateral $MAOB$, where $O$ is the origin, is

Statement $(A) :$ For all values of $\theta$,the line $(x - 3) \cos \theta + (y - 3) \sin \theta = 1$ is tangent to the circle $(x - 3)^2 + (y - 3)^2 = 1$.
Reason $(R) :$ For all values of $\theta$,the line $x \cos \theta + y \sin \theta = a$ is tangent to the circle $x^2 + y^2 = a^2$.