## Precalculus (6th Edition)

center: $(2, -4)$ radius = $\sqrt{17}$
RECALL: (1) The graph of the equation $(x-h)^2+(y-k)^2=r^2$ is a circle whose center is at $(h, k)$ and whose radius is $r$ units. (2) To complete the square for the binomial $x^2+bx$, $(\frac{b}{2})^2$ must be added. The factored form of the trinomial is $(x+\frac{b}{2})^2$. Subtract $3$ to both sides of the equation to obtain: $$x^2+y^2-4x+8y=-3$$ Group the terms with the same variable to obtain: $$(x^2-4x)+(y^2+8y)=-3$$ Complete the square for each binomial using the formula in (2) above. Make sure to add the same quantity added on the left side of the equation to the right side of the equation to maintain the equality of both sides. $$(x^2-4x+\color{red}{(\frac{-4}{2})^2})+(y^2+8y+\color{blue}{(\frac{8}{2})^2})=-3+\color{red}{(\frac{-4}{2})^2)}+\color{blue}{(\frac{8}{2})^2} \\(x^2-4x+4)+(y^2+8y+16)=-3+4+16 \\(x-2)^2+(y+4)^2=17$$ The graph of the equation above is a circle whose center is at $(2, -4)$ and has $r^2=17$. Thus, $r=\sqrt{17}$. Therefore, the given circle has: center: $(2, -4)$ radius = $\sqrt{17}$