Find the equation of the pair of tangents drawn from the origin to the circle $x^2 + y^2 + 20(x + y) + 20 = 0$.

If the shortest distance between the parabola $y^2=4x$ and the circle $x^2+y^2-4x-16y+64=0$ is $d$,then $d^2$ is equal to:

$A$ tangent $PT$ is drawn to the circle $x^2+y^2=4$ at the point $P(\sqrt{3}, 1)$. $A$ straight line $L$,perpendicular to $PT$,is a tangent to the circle $(x-3)^2+y^2=1$.
$1.$ $A$ common tangent of the two circles is
$(A)$ $x=4$ $(B)$ $y=2$ $(C)$ $x+\sqrt{3} y=4$ $(D)$ $x+2 \sqrt{2} y=6$
$2.$ $A$ possible equation of $L$ is
$(A)$ $x-\sqrt{3} y=1$ $(B)$ $x+\sqrt{3} y=1$ $(C)$ $x-\sqrt{3} y=-1$ $(D)$ $x+\sqrt{3} y=5$

Two chords of the circle $x^2+y^2-2gx-2hy+g^2+h^2-c^2=0$ pass through the point $(g, h+c)$,and the line $y=x$ bisects these two chords. Then:

The centre of the circle passing through the point $(0,1)$ and touching the parabola $y=x^{2}$ at the point $(2,4)$ is

If the line $x \cos \theta + y \sin \theta = 2$ is the equation of a transverse common tangent to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 6 \sqrt{3} x - 6y + 20 = 0$,then the value of $\theta$ is: