The circles $x^2+y^2-2x-4y-4=0$ and $x^2+y^2+2x+4y-11=0$:

The radius of the circle,having centre at $(2, 1)$ and one of its chords as a diameter of the circle $x^2 + y^2 - 2x - 6y + 6 = 0$,is

The length of the chord joining the points in which the straight line $\frac{x}{3} + \frac{y}{4} = 1$ cuts the circle ${x^2} + {y^2} = \frac{169}{25}$ is

$A$ circle with centre $(2, 3)$ and radius $4$ intersects the line $x + y = 3$ at the points $P$ and $Q$. If the tangents at $P$ and $Q$ intersect at the point $S(\alpha, \beta)$,then $4 \alpha - 7 \beta$ is equal to $........$.

The equation of the pair of tangents at $(0,1)$ to the circle $x^{2}+y^{2}-2x-6y+6=0$ is

If the circle $S_1: x^2+y^2=16$ intersects another circle $S_2$ of radius $5$ units such that the common chord is of maximum length and has a slope of $\frac{3}{4}$,then the center of the circle $S_2$ is