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

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 $2.$

If $OA$ and $OB$ be the tangents to the circle ${x^2} + {y^2} – 6x – 8y + 21 = 0$ drawn from the origin $O$, then $AB =$

The line $lx + my + n = 0$ is normal to the circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$, if

The angle at which the circles $(x – 1)^2 + y^2 = 10$ and $x^2 + (y – 2)^2 = 5$ intersect is