Let the lines $y+2 x=\sqrt{11}+7 \sqrt{7}$ and $2 y + x =2 \sqrt{11}+6 \sqrt{7}$ be normal to a circle $C:(x-h)^{2}+(y-k)^{2}=r^{2}$. If the line $\sqrt{11} y -3 x =\frac{5 \sqrt{77}}{3}+11$ is tangent to the circle $C$, then the value of $(5 h-8 k)^{2}+5 r^{2}$ is equal to…….

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

If the line $3x + 4y – 1 = 0$ touches the circle ${(x – 1)^2} + {(y – 2)^2} = {r^2}$, then the value of $r$ will be

The equations of the tangents to the circle ${x^2} + {y^2} = {a^2}$ parallel to the line $\sqrt 3 x + y + 3 = 0$ are

The tangent$(s)$ from the point of intersection of the lines $2x -3y + 1$ = $0$ and $3x -2y -1$ = $0$ to circle $x^2 + y^2 + 2x -4y$ = $0$ will be –