The Center of the Circle $5(x-7)^{2}+4(y+4)^{2}-125=-(y+4)^{2}$ Is At $(\ldots, \ldots)$

Short Answer

The center of the circle $5(x-7)^{2}+4(y+4)^{2}-125=-(y+4)^{2}$ is at $(\ldots, \ldots)$ .