$A$ circle is inscribed in an equilateral triangle of side length $6$. Find the area of the square inscribed in this circle.

Three circles,each of radius $1$,touch one another externally and lie between two parallel lines. The minimum possible distance between the lines is:

$A$ square is inscribed in the circle $x^2 + y^2 - 6x + 8y - 103 = 0$ with its sides parallel to the coordinate axes. Then the distance of the vertex of the square which is nearest to the origin is

The straight line $2x - 3y = 1$ divides the circular region $x^2 + y^2 \leq 6$ into two parts. If $S = \left\{ \left(2, \frac{3}{4}\right), \left(\frac{5}{2}, \frac{3}{4}\right), \left(\frac{1}{4}, -\frac{1}{4}\right), \left(\frac{1}{8}, \frac{1}{4}\right) \right\}$,then the number of point$(s)$ in $S$ lying inside the smaller part is

Choose the incorrect statement about the two circles whose equations are given below:
$x^{2}+y^{2}-10x-10y+41=0$ and $x^{2}+y^{2}-16x-10y+80=0$

$A$ line is drawn through a fixed point $P(\alpha, \beta)$ to cut the circle $x^{2}+y^{2}=r^{2}$ at $A$ and $B$. Then $PA \cdot PB$ is equal to