If a circle $S$ with radius $5$ touches the circle $x^2+y^2-6x-4y-12=0$ at $(-1,-1)$,then the length of the tangent from the centre of the circle $S$ to the given circle 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.

The number of common tangents to the circles $x^2+y^2-2x-6y+9=0$ and $x^2+y^2+6x-2y+1=0$ is

Let a circle $C_1 \equiv x^2 + y^2 - 4x + 6y + 1 = 0$ and circle $C_2$ be such that its centre is the image of the centre of $C_1$ about the $x$-axis and the radius of $C_2$ is equal to the radius of $C_1$. Then,the area of $C_1$ which is not common with $C_2$ 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:

Let the equation of the circle,which touches the $x$-axis at the point $(a, 0), a > 0$ and cuts off an intercept of length $b$ on the $y$-axis,be $x^2 + y^2 - \alpha x + \beta y + \gamma = 0$. If the circle lies below the $x$-axis,then the ordered pair $(2a, b^2)$ is equal to: