The equation of the normal to the circle ${x^2} + {y^2} = 9$ at the point $\left( {\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right)$ is
$x + y = 0$
$x - y = \frac{{\sqrt 2 }}{3}$
$x - y = 0$
None of these
If $\frac{x}{\alpha } + \frac{y}{\beta } = 1$ touches the circle ${x^2} + {y^2} = {a^2}$, then point $(1/\alpha ,\,1/\beta )$ lies on a/an
The point of contact of the tangent to the circle ${x^2} + {y^2} = 5$ at the point $(1, -2)$ which touches the circle ${x^2} + {y^2} - 8x + 6y + 20 = 0$, is
Tangents are drawn from the point $(4, 3)$ to the circle ${x^2} + {y^2} = 9$. The area of the triangle formed by them and the line joining their points of contact 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 $2.$