If the equation of the incircle of an equilateral triangle is $x^2 + y^2 + 4x - 6y + 4 = 0$,then the equation of its circumcircle is:

Let $x-y=0$ and $x+y=1$ be two perpendicular diameters of a circle of radius $R$. The circle will pass through the origin if $R$ is equal to

The interval of the values of $a$ for which the line $x + y = 0$ bisects $2$ distinct chords drawn from a point $P \left( \frac{1 + \sqrt{2} a}{2}, \frac{1 - \sqrt{2} a}{2} \right)$ to the circle $2x^2 + 2y^2 - (1 + \sqrt{2} a)x - (1 - \sqrt{2} a)y = 0$ is:

Let $PQ$ and $RS$ be the tangents at the extremities of the diameter $PR$ of a circle of radius $r$. If $PS$ and $RQ$ intersect at a point $X$ on the circumference of the circle,then $(PQ \cdot RS)$ is equal to

$A$ circle touches both the coordinate axes and the straight line $L \equiv 4x+3y-6=0$ in the first quadrant. If this circle lies below the line $L=0$,then the equation of that circle is

The area (in square units) of the circle which touches the lines $4x + 3y = 15$ and $4x + 3y = 5$ is