The line $x + y = 2$ is tangent to the curve $x^2 = 3 - 2y$ at its point

Statement $(A) :$ For all values of $\theta$,the line $(x - 3) \cos \theta + (y - 3) \sin \theta = 1$ is tangent to the circle $(x - 3)^2 + (y - 3)^2 = 1$.
Reason $(R) :$ For all values of $\theta$,the line $x \cos \theta + y \sin \theta = a$ is tangent to the circle $x^2 + y^2 = a^2$.

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

Let the lengths of intercepts on the $x$-axis and $y$-axis made by the circle $x^{2}+y^{2}+ax+2ay+c=0$ $(a < 0)$ be $2\sqrt{2}$ and $2\sqrt{5}$,respectively. Then the shortest distance from the origin to a tangent to this circle which is perpendicular to the line $x+2y=0$ is equal to:

If the line $4x - 3y + p = 0$ $(p + 3 > 0)$ touches the circle $x^2 + y^2 - 4x + 6y + 4 = 0$ at the point $(h, k)$,then $h - 2k = . . . . . .$

Find the equation of the normal to the circle $x^2 + y^2 - 5x + 2y - 48 = 0$ at the point $(5, 6)$.