The tangent to the circle $x^2+y^2=5$ at $(1,-2)$ also touches the circle $x^2+y^2-8x+6y+20=0$. The coordinates of the corresponding point of contact are:

The point on the curve $x^2+y^2=a^2, y \geq 0$,at which the tangent is parallel to the $x$-axis is

$A$ circle passes through the points $(-1, 1)$,$(0, 6)$,and $(5, 5)$. The point$(s)$ on this circle,the tangent$(s)$ at which is/are parallel to the straight line joining the origin to its centre is/are:

The equation of the tangent to the circle $x^2 + y^2 = a^2$ which makes a triangle of area $a^2$ with the coordinate axes is:

Consider a circle $(x-\alpha)^2+(y-\beta)^2=50$,where $\alpha, \beta > 0$. If the circle touches the line $y+x=0$ at the point $P$,whose distance from the origin is $4 \sqrt{2}$,then $(\alpha+\beta)^2$ is equal to................

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