The equation of tangents to the circle $x^2+y^2=4$ which are parallel to $x+2y+3=0$ are

The equation of the tangent to the circle,given by $x=5 \cos \theta, y=5 \sin \theta$ at the point $\theta=\frac{\pi}{3}$ on it,is

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

The normal to the circle $x^2 + y^2 - 3x - 6y - 10 = 0$ at the point $(-3, 4)$ is

The equation of the tangent to the circle $x^2 + y^2 = r^2$ at the point $(a, b)$ is $ax + by - \lambda = 0$,where $\lambda$ is:

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