The slope of the tangent to the circle $(x-6)^2 + y^2 = 2$,which passes through the focus of the parabola $y^2 = 16x$,is:

The line $x \cos \alpha + y \sin \alpha = p$ will be a tangent to the circle $x^2 + y^2 - 2ax \cos \alpha - 2ay \sin \alpha = 0$,if $p = $

The point on the curve ${y^2} = 2(x - 3)$ at which the normal is parallel to the line $y - 2x + 1 = 0$ 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:

If a line,$y=mx+c$ is a tangent to the circle,$(x-3)^{2}+y^{2}=1$ and it is perpendicular to a line $L_{1},$ where $L_{1}$ is the tangent to the circle,$x^{2}+y^{2}=1$ at the point $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right),$ then

The equations of the tangents drawn from the origin to the circle $x^2+y^2+2gx+2fy+g^2=0$ are