Let the equation of a circle be $x^{2}+y^{2}-6x-4y+9=0$. Then the line $4x+3y-8=0$ is a

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 angle between the two tangents from the origin to the circle $(x - 7)^2 + (y + 1)^2 = 25$ is

$A$ circle $S=0$ with radius $\sqrt{2}$ touches the line $x+y-2=0$ at $(1,1)$. Then,the length of the tangent drawn from the point $(1,2)$ to $S=0$ is

The straight line $x+2y=1$ cuts the $X$-axis at $A$ and the $Y$-axis at $B$. $A$ circle is drawn through $A, B$ and the origin $O(0,0)$. The sum of the perpendicular distances from $A$ and $B$ to the tangent drawn at the origin to the circle $S$ is:

The equation of a circle which has a tangent $3x + 4y = 6$ and two normals given by $(x - 1)(y - 2) = 0$ is