The length of the shortest path that begins at the point $(2, 5)$,touches the $x-$axis,and then ends at a point on the circle $x^2 + y^2 + 12x - 20y + 120 = 0$.

The number of common tangents to the circles $x^2+y^2-18x-15y+131=0$ and $x^2+y^2-6x-6y-7=0$ is:

Let $6$ and $8$ be the $X$ and $Y$-intercepts made by the circle $S \equiv x^2+y^2+2gx+2fy+c=0$ respectively. If $gx+fy+1=0$ is a line passing through the point $(1, -1)$,then the radius of the circle $S=0$ is

Two circles in the first quadrant of radii $r_1$ and $r_2$ touch the coordinate axes. Each of them cuts off an intercept of $2$ units with the line $x+y=2$. Then $r_1^2+r_2^2-r_1 r_2$ is equal to $...........$

If the length of the chord $2x + 3y + k = 0$ of the circle $x^2 + y^2 - 2x + 4y - 11 = 0$ is $2\sqrt{3}$,then the sum of all possible values of $k$ is

The point $(5, -7)$ lies outside the circle: