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

If the line $ax+by=1$ is a tangent to the circle $S_r \equiv x^2+y^2-r^2=0$,then which one of the following is true?

If one common tangent of the two circles $x^2 + y^2 = 4$ and $x^2 + (y - 3)^2 = \lambda, \lambda > 0$ passes through the point $(\sqrt{3}, 1)$,then the possible value of $\lambda$ 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 $x^2 + y^2 - 22x - 4y + 25 = 0$ which is perpendicular to the line $5x + 12y + 8 = 0$ 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: