A tangent $P T$ is drawn to the circle $x^2+y^2=4$ at the point $P(\sqrt{3}, 1)$. A straight line $L$, perpendicular to $P T$ 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

$(A)$ $x-\sqrt{3} y=1$ $(B)$ $x+\sqrt{3} y=1$ $(C)$ $x-\sqrt{3} y=-1$ $(D)$ $x+\sqrt{3} y=5$

Give the answer question $1$ and $2.$

$x = 7$ touches the circle ${x^2} + {y^2} - 4x - 6y - 12 = 0$, then the coordinates of the point of contact are

If $\theta $ is the angle subtended at $P({x_1},{y_1})$ by the circle $S \equiv {x^2} + {y^2} + 2gx + 2fy + c = 0$, then

Two concentric circles are such that the smaller divides the larger into two regions of equal area. If the radius of the smaller circle is $2$ , then the length of the tangent from any point $' P '$ on the larger circle to the smaller circle is :

Let the tan gents drawn to the circle, $x^2 + y^2 = 16$ from the point $P(0, h)$ meet the $x-$ axis at point  $A$ and $B.$ If the area of $\Delta APB$ is minimum, then $h$  is equal to

The area of triangle formed by the tangent, normal drawn at $(1,\sqrt 3 )$ to the circle ${x^2} + {y^2} = 4$ and positive $x$-axis, is