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

Tangents drawn from the point $P(1,8)$ to the circle $x^2+y^2-6 x-4 y-11=0$ touch the circle at the points $A$ and $B$. The equation of the circumcircle of the triangle $P A B$ is

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 value of $c$, for which the line $y = 2x + c$ is a tangent to the circle ${x^2} + {y^2} = 16$, is

The line $3x - 2y = k$ meets the circle ${x^2} + {y^2} = 4{r^2}$ at only one point, if ${k^2}$=

A circle $C_{1}$ passes through the origin $O$ and has diameter $4$ on the positive $x$-axis. The line $y =2 x$ gives a chord $OA$ of a circle $C _{1}$. Let $C _{2}$ be the circle with $OA$ as a diameter. If the tangent to $C _{2}$ at the point $A$ meets the $x$-axis at $P$ and $y$-axis at $Q$, then $QA : AP$ is equal to.