Given the circles ${x^2} + {y^2} - 4x - 5 = 0$and ${x^2} + {y^2} + 6x - 2y + 6 = 0$. Let $P$ be a point $(\alpha ,\beta )$such that the tangents from P to both the circles are equal, then

The normal at the point $(3, 4)$ on a circle cuts the circle at the point $(-1, -2)$. Then the equation of the circle is

The equation of circle which touches the axes of coordinates and the line $\frac{x}{3} + \frac{y}{4} = 1$ and whose centre lies in the first quadrant is ${x^2} + {y^2} - 2cx - 2cy + {c^2} = 0$, where $c$ is

The equations of tangents to the circle ${x^2} + {y^2} - 22x - 4y + 25 = 0$ which are perpendicular to the line $5x + 12y + 8 = 0$ are

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 two circles which passes through $(0,a)$ and $(0, - a)$ and touch the line $y = mx + c$ will intersect each other at right angle, if