Consider circle $S$ : $x^2 + y^2 = 1$ and $P(0, -1)$ on it. $A$ ray of light gets reflected from tangent to $S$ at $P$ from the point with abscissa $-3$ and becomes tangent to the circle $S.$  Equation of reflected ray is

$x = 7$ touches the circle ${x^2} + {y^2} - 4x - 6y - 12 = 0$, then the coordinates of the point of contact are

Let $A B$ be a chord of length $12$ of the circle $(x-2)^{2}+(y+1)^{2}=\frac{169}{4}$ If tangents drawn to the circle at points $A$ and $B$ intersect at the point $P$, then five times the distance of point $P$ from chord $AB$ 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 $2.$

Two circles each of radius $5\, units$ touch each other at the point $(1,2)$. If the equation of their common tangent is $4 \mathrm{x}+3 \mathrm{y}=10$, and $\mathrm{C}_{1}(\alpha, \beta)$ and $\mathrm{C}_{2}(\gamma, \delta)$, $\mathrm{C}_{1} \neq \mathrm{C}_{2}$ are their centres, then $|(\alpha+\beta)(\gamma+\delta)|$ is equal to .... .

The line $y = x + c$will intersect the circle ${x^2} + {y^2} = 1$ in two coincident points, if