The radius of a circle whose center lies in the fourth quadrant and touches each of the three lines $x=0$,$y=0$,and $3x+4y-12=0$ is .... units.

Let $C: x^2+y^2=4$ and $C^{\prime}: x^2+y^2-4 \lambda x+9=0$ be two circles. If the set of all values of $\lambda$ such that the circles $C$ and $C^{\prime}$ intersect at two distinct points is $\mathbb{R}-[a, b]$,then the point $(8a+12, 16b-20)$ lies on the curve:

Let the centre of the circle be in the first quadrant and lie on the line $2x - y = 4$. Let the area of an equilateral triangle inscribed in the circle be $27\sqrt{3}$. Then the square of the length of the chord of the circle on the line $x = 1$ is . . . . . .

Let the mirror image of a circle $c_{1}: x^{2}+y^{2}-2x-6y+\alpha=0$ in the line $y=x+1$ be $c_{2}: 5x^{2}+5y^{2}+10gx+10fy+38=0$. If $r$ is the radius of circle $c_{2}$,then $\alpha+6r^{2}$ is equal to:

If the ratio of the lengths of the tangents drawn from the point $(1, 2)$ to the circles $x^2 + y^2 + x + y - 4 = 0$ and $3x^2 + 3y^2 - x - y + k = 0$ is $4 : 3$,then $k = \dots$

The position of the point $(0.1, 3.1)$ with respect to the circle $x^2 + y^2 - 2x - 4y + 3 = 0$ is: