The equation of the circle inscribed in a square formed by the lines $x+y-2=0$,$x+y-6=0$,$x-y+1=0$ and $x-y+5=0$ is

The equation of the circle which touches the circle $S \equiv x^2+y^2-10x-4y+19=0$ at the point $(2,3)$ internally and has a radius equal to half of the radius of the circle $S=0$ is:

For the circle $x-2=5 \cos \theta, y+1=5 \sin \theta$ where $\theta$ is the parameter,the line $x=1+\frac{r}{2}, y=-2+\frac{\sqrt{3}}{2} r$ where $r$ is the parameter,is a

$A$ circle touching the $x-$ axis at $(3, 0)$ and making an intercept of length $8$ on the $y-$ axis passes through the point:

The equation of a circle touching the curve $|x - 1| + |y - 4| = 6$ is

Let $C$ be the centre of the circle $x^{2}+y^{2}-x+2 y=\frac{11}{4}$ and $P$ be a point on the circle. $A$ line passes through the point $C$,makes an angle of $\frac{\pi}{4}$ with the line $CP$ and intersects the circle at the points $Q$ and $R$. Then the area of the triangle $PQR$ (in unit$^{2}$) is.