The slope of the common tangent drawn to the circles $x^2+y^2-4x+12y-216=0$ and $x^2+y^2+6x-12y+36=0$ is

$A$ triangle is formed by the tangents at the point $(2,2)$ on the curves $y^2=2x$ and $x^2+y^2=4x$,and the line $x+y+2=0$. If $r$ is the radius of its circumcircle,then $r^2$ is equal to $........$.

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

If $y=\sqrt{3}x+k_1$ and $y=\sqrt{3}x+k_2$ are two parallel tangents of a circle of radius $2 \text{ units}$,then $|k_1-k_2|$ is equal to

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

The angle between the two tangents drawn from the point $(\alpha, \beta)$ to the circle $x^{2} + y^{2} = a^{2}$ is: