The equation of the normal to the circle $x^2 + y^2 = 9$ at the point $\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$ is

If the tangents $x+y+k=0$ and $x+ay+b=0$ drawn to the circle $S \equiv x^2+y^2+2x-2y+1=0$ are perpendicular to each other and $k, b$ are both greater than $1$,then $b-k=$

The lines $y - y_1 = m(x - x_1) \pm a \sqrt{1 + m^2}$ are tangents to the same circle. The radius of the circle is:

The equation of a normal to the circle $x^2+y^2-2x=0$ that is parallel to the line $x+2y-3=0$ 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 the pair of tangents drawn to the circle $x^2+y^2=a^2$ from the point $(10,4)$ are perpendicular,then $a=$