If the tangents drawn at the point $O (0,0)$ and $P (1+\sqrt{5}, 2)$ on the circle $x ^{2}+ y ^{2}-2 x -4 y =0$ intersect at the point $Q$, then the area of the triangle $OPQ$ is equal to

If the ratio of the lengths of tangents drawn from the point $(f,g)$ to the given circle ${x^2} + {y^2} = 6$ and ${x^2} + {y^2} + 3x + 3y = 0$ be $2 : 1$, then

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

If the area of the triangle formed by the positive $x-$axis, the normal and the tangent to the circle $(x-2)^{2}+(y-3)^{2}=25$ at the point $(5,7)$ is $A$ then $24 A$ is equal to ...... .

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 $2.$

A tangent to the circle ${x^2} + {y^2} = 5$at the point $(1,-2)$ the circle ${x^2} + {y^2} - 8x + 6y + 20 = 0$