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 $2.$

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

The line $2x - y + 1 = 0$ is tangent to the circle at the point $(2, 5)$ and the centre of the circles lies on $x-2y=4$. The radius of the circle is 

Tangents are drawn from the point $(-1,-4)$ to the circle $x^2 + y^2 - 2x + 4y + 1 = 0$. Length of corresponding chord of contact will be-

In the given figure, $AB$ is tangent to the circle with centre $O$ , the ratio of the shaded region to the unshaded region of the triangle $OAB$ is

The equation of normal to the circle $2{x^2} + 2{y^2} - 2x - 5y + 3 = 0$ at $(1, 1)$ is