The Cartesian equation of the curve given by $x = 6 \cos \theta$ and $y = 6 \sin \theta$ is

Let a circle $C$ have its centre in the first quadrant, intersect the coordinate axes at exactly three points and cut off equal intercepts from the coordinate axes. If the length of the chord of $C$ on the line $x + y = 1$ is $\sqrt{14}$, then the square of the radius of $C$ is . . . . . .

$P$ and $Q$ are the points of trisection of the line segment joining the points $(3, -7)$ and $(-5, 3)$. If $PQ$ subtends a right angle at a variable point $R$,then the locus of $R$ is

The centre of the circle passing through the point $(0,1)$ and touching the curve $y=x^2$ at $(2,4)$ is

The equation of the circle which touches the $x$-axis at $(3, 0)$ and passes through $(1, 4)$ is given by

The equation of the circle,the end-points of whose diameter are the centres of the circles $x^{2}+y^{2}-2x+3y-3=0$ and $x^{2}+y^{2}+6x-12y-5=0$ is