At which point does the line $x = 0$ touch the circle $x^2 + y^2 - 2x - 6y + 9 = 0$?

The circle $x^2+y^2-4x-8y+16=0$ rolls along the tangent drawn to it at $(2+\sqrt{3}, 3)$ by $2$ units. The equation of the circle in the new position is

Consider the following statements:
Assertion $(A)$: The circle $x^2 + y^2 = 1$ has exactly two tangents parallel to the $x$-axis.
Reason $(R)$: $\frac{dy}{dx} = 0$ on the circle exactly at the points $(0, \pm 1)$.
Of these statements:

If $2x - 3y + 5 = 0$ and $4x - 5y + 7 = 0$ are the equations of the normals drawn to a circle and $(2, 5)$ is a point on the given circle,then the radius of the circle is

The equations of the normals to the circle $x^2 + y^2 - 8x - 2y + 12 = 0$ at the points whose ordinate is $-1$ are:

The normal to the circle given by $x^2+y^2-6x+8y-144=0$ at $(8,8)$ meets the circle again at the point