The straight line $x - y - 3 = 0$ touches the circle $x^2 + y^2 - 4x + 6y + 11 = 0$ at the point whose coordinates are

The equation of the tangent to the circle $x^2 + y^2 = 25$ which is inclined at an angle of $60^{\circ}$ with the $x$-axis is:

$A$ circle $C_{1}$ passes through the origin $O$ and has a diameter of $4$ on the positive $x$-axis. The line $y = 2x$ intersects the circle $C_{1}$ at $O$ and $A$. Let $C_{2}$ be the circle with $OA$ as a diameter. If the tangent to $C_{2}$ at the point $A$ meets the $x$-axis at $P$ and the $y$-axis at $Q$,then the ratio $QA : AP$ is equal to:

The lines $y - y_1 = m(x - x_1) \pm a \sqrt{1 + m^2}$ are tangents to the same circle. The radius of the circle is:

If the tangents $x+y+k=0$ and $x+ay+b=0$ drawn to the circle $S \equiv x^2+y^2+2x-2y+1=0$ are perpendicular to each other and $k, b$ are both greater than $1$,then $b-k=$

If the lines $3x + 4y - 14 = 0$ and $6x + 8y + 7 = 0$ are both tangents to a circle,then its radius is