The equation of the circle which touches the coordinate axes and the line $\frac{x}{3} + \frac{y}{4} = 1$,and whose centre lies in the first quadrant,is ${x^2} + {y^2} - 2cx - 2cy + {c^2} = 0$,where $c$ 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.

Let $S=0$ be the circle passing through the points $(2,0)$,$(1,-2)$,and $(-1,1)$. Then the point $(1,2)$

$A$ square is inscribed in the circle $x^2+y^2-2x+4y-93=0$ with its sides parallel to the coordinate axes. Which of the following can be one of the vertices of the square?

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:

Let $x_0, y_0$ be fixed real numbers such that $x_0^2+y_0^2 > 1$. If $x, y$ are arbitrary real numbers such that $x^2+y^2 \leq 1$,then the minimum value of $(x-x_0)^2+(y-y_0)^2$ is