The equation of pair of tangents to the circle ${x^2} + {y^2} - 2x + 4y + 3 = 0$ from $(6, - 5)$, is

The angle between the pair of tangents from the point $(1, 1/2)$ to the circle $x^2 + y^2 + 4x + 2y -4=0$ is-

A tangent to the circle ${x^2} + {y^2} = 5$at the point $(1,-2)$ the circle ${x^2} + {y^2} - 8x + 6y + 20 = 0$

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

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

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.......