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

At which point on the $y$-axis is the line $x = 0$ a tangent to the circle $x^2 + y^2 - 2x - 6y + 9 = 0$?

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 area (in sq. units) of the triangle formed by the $X$-axis,the tangent,and the normal drawn to the circle $x^2+y^2=10x$ at the point $(9,3)$ is

If the line $x+3y=0$ is the tangent at $(0,0)$ to the circle of radius $1$,then the centre of one such circle 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 = $