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

If $\Delta$ is the area of the triangle formed by the positive $x$-axis and the normal and tangent to the circle $x^2+y^2=4$ at $(1, \sqrt{3})$,then $\Delta$ is equal to

Let the lengths of intercepts on the $x$-axis and $y$-axis made by the circle $x^{2}+y^{2}+ax+2ay+c=0$ $(a < 0)$ be $2\sqrt{2}$ and $2\sqrt{5}$,respectively. Then the shortest distance from the origin to a tangent to this circle which is perpendicular to the line $x+2y=0$ is equal to:

The product of the slopes of the common tangents drawn to the circles $x^2+y^2+2x-2y-2=0$ and $x^2+y^2-2x+2y+1=0$ is:

Point $M$ moves along the circle $(x - 4)^2 + (y - 8)^2 = 20$. It then breaks away from the circle and moves along a tangent to the circle,passing through the point $(-2, 0)$ on the $x$-axis. The coordinates of the point on the circle at which the moving point broke away can be: