If $2x - 4y = 9$ and $6x - 12y + 7 = 0$ are the tangents of same circle, then its radius will be
$\frac{{\sqrt 3 }}{5}$
$\frac{{17}}{{6\sqrt 5 }}$
$\frac{{2\sqrt 5 }}{3}$
$\frac{{17}}{{3\sqrt 5 }}$
The tangent at $P$, any point on the circle ${x^2} + {y^2} = 4$, meets the coordinate axes in $A$ and $B$, then
The slope of the tangent at the point $(h,h)$ of the circle ${x^2} + {y^2} = {a^2}$ is
Square of the length of the tangent drawn from the point $(\alpha ,\beta )$ to the circle $a{x^2} + a{y^2} = {r^2}$ is
Let a circle $C$ of radius $5$ lie below the $x$-axis. The line $L_{1}=4 x+3 y-2$ passes through the centre $P$ of the circle $C$ and intersects the line $L _{2}: 3 x -4 y -11=0$ at $Q$. The line $L _{2}$ touches $C$ at the point $Q$. Then the distance of $P$ from the line $5 x-12 y+51=0$ is
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