The normal to the circle given by $x^2+y^2-6x+8y-144=0$ at $(8,8)$ meets the circle again at the point

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 $x^2 + y^2 = a^2$.

The line $lx + my + n = 0$ will be a tangent to the circle $x^2 + y^2 = a^2$ if

Let $A$ be the centre of the circle $x^2+y^2-2x-4y-20=0$. If the tangents drawn at the points $B(1,7)$ and $D(4,-2)$ on the given circle meet at the point $C$,then the area of the quadrilateral $ABCD$ is

Which of the following lines is a tangent to the circle $x^2 + y^2 = 25$ for all values of $m$?

If the length of the tangent drawn from the point $(5, 3)$ to the circle $x^2 + y^2 + 2x + ky + 17 = 0$ is $7$,then the value of $k$ is: