The length of the chord intercepted by the circle $x^2+y^2-8x-2y-8=0$ on the line $x+y+1=0$ is:

Let $B$ be the centre of the circle $x^{2}+y^{2}-2x+4y+1=0$. Let the tangents at two points $P$ and $Q$ on the circle intersect at the point $A(3,1)$. Then $8 \left(\frac{\text{Area } \triangle APQ}{\text{Area } \triangle BPQ}\right)$ is equal to:

If the coordinates of one end of a diameter of the circle $x^2+y^2+4x-8y+5=0$ are $(2,1)$, then the coordinates of the other end are:

$A$ circle passes through the point $(0, 1)$ and touches the parabola $y = x^2$ at the point $(1, 1)$. The centre of the circle is...

$A$ line $l$ meets the circle $x^2+y^2=61$ at points $A$ and $B$. If $P(-5, 6)$ is a point such that $PA=PB=10$,then the equation of line $l$ is:

Statement $(A):$ The number of common tangents to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 6x - 8y = 24$ is $4$.
Reason $(R):$ For two circles with centers $C_1, C_2$ and radii $r_1, r_2$,if $|C_1C_2| > r_1 + r_2$,then the circles have $4$ common tangents.