The equation of the circle whose end points of a diameter are the centres of the circles $x^{2}+y^{2}+2x-4y+1=0$ and $x^{2}+y^{2}-8x+6y+17=0$ is

If the equation of the circle lying in the first quadrant,touching both the coordinate axes and the line $\frac{x}{3}+\frac{y}{4}=1$ is $(x-c)^2+(y-c)^2=c^2$,then $c=$

If ${g^2} + {f^2} = c$,then the equation ${x^2} + {y^2} + 2gx + 2fy + c = 0$ will represent

Find the equation of the circle with centre $(1, 1)$ and radius $\sqrt{2}$.

The equation of the circle which touches the lines $x = 0$,$y = 0$ and $3x + 4y = 4$ is

The equation of a circle is $\operatorname{Re}(z^{2})+2(\operatorname{Im}(z))^{2}+2 \operatorname{Re}(z)=0$,where $z=x+iy$. $A$ line which passes through the center of the given circle and the vertex of the parabola $x^{2}-6x-y+13=0$ has a $y$-intercept equal to $.....$