## Precalculus (6th Edition)

The graph of the given equation is a circle with: center at $(-\frac{1}{2}, 2)$; a radius of $3$
RECALL: (1) The center-radius form of a circle's equation is $(x-h)^2+(y-k)^2=r^2$ with center at $(h, k)$ and a radius of $r$ units. (2) The radius of a circle must be positive. (3) To complete the square for $x^2+bx$, $(\frac{b}{2})^2$ must be added. After completing the square, the factored form of the perfect square trinomial is $(x+b)^2$. Add $19$ to both sides of the equation to obtain: $4x^2+4y^2+4x-16y=19$ Divide both sides by $4$ to obtain: $\dfrac{4x^2+4y^2+4x-16y}{4}=\dfrac{19}{4} \\x^2+y^2+x-4y=\frac{19}{4}$ Group terms with the same variables to obtain: $(x^2+x)+(y^2-4y)=\frac{19}{4}$ Complete the square for each binomial using the formula in (3) above. Make sure to add to the right side of the equation whatever is added on the left side to maintain the equality of both sides. $(x^2+x+\color{red}{(\frac{1}{2})^2}) + (y^2-4y+\color{red}{(\frac{-4}{2})^2}) = \frac{19}{4} +\color{red}{(\frac{1}{2})^2+(\frac{-4}{2})^2} \\(x^2+x+\frac{1}{4})+(y^2-4y+4)=\frac{19}{4}+\frac{1}{4}+4 \\(x^2+x+\frac{1}{4})+(y^2-4y+4)=\frac{20}{4}+4 \\(x+\frac{1}{2})^2+(y-2)^2=9 \\(x+\frac{1}{2})^2+(y-2)^2=3^2$ This equation is in the same form as the formula/equation in (1) above. Thus, its graph is a circle with center at $(-\frac{1}{2}, 2)$ and a radius of $3$.