Respuesta :
We have the following:
[tex]36x^2+36y^2-36x+48y=-16[/tex]The equation of a circle when the center is not the origin is as follows
[tex]\begin{gathered} (x-h)^2+(y-k)^2=r^2 \\ (h,k)\rightarrow center \\ r\rightarrow radius \end{gathered}[/tex]we organize and solve
[tex]\begin{gathered} \frac{36}{36}x^2+\frac{36}{36}y^2-\frac{36}{36}x+\frac{48}{36}y=-\frac{16}{36} \\ x^2+y^2-x+\frac{4}{3}y=-\frac{4}{9} \\ (x^2-x+\frac{1}{4})+(y^2+\frac{4}{3}y+\frac{4}{9})=-\frac{16}{36}+\frac{1}{4}+\frac{4}{9} \\ (x-\frac{1}{2})^2+(y+\frac{2}{3})^2=\frac{1}{4} \\ (x-\frac{1}{2})^2+(y+\frac{2}{3})^2=(\frac{1}{2})^2 \\ \text{center}=(\frac{1}{2},-\frac{2}{3}) \\ \text{radius}=\frac{1}{2} \end{gathered}[/tex]Therefore, in this case if we have a circle with center at (1/2, -2/3) and the one with a radius of 1/2
