## Precalculus (6th Edition)

Published by Pearson

# Chapter 2 - Graphs and Functions - Chapter 2 Test Prep - Review Exercises - Page 297: 17

#### Answer

center at $(-\frac{7}{2}, -\frac{3}{2})$ radius = $\frac{5\sqrt2}{2}$ units

#### Work Step by Step

RECALL: (1) The center-radius form of a circle's equation is $(x-h)^2+(y-k)^2=r^2$ where $(h, k)$ is the center and $r$ = radius (2) To complete the square for $x^2+b^2$, add $(\frac{b}{2})^2$. The factored form of the perfect square trinomial is $(x+\frac{b}{2})^2$. To write the given equation in center-radius form, perform the following steps: (1) Divide $2$ to both sides of the equation: $$\dfrac{2x^2+2y^2+14x+6y+2}{2}=\dfrac{0}{2} \\x^2+y^2+7x+3y+1=0$$ (2) Subtract $1$ to both sides of the equation. $$x^2+y^2+7x+3y=-1$$ (3) Group terms terms with the same variable: $$(x^2+7x)+(y^2+3y)=-1$$ (3) Complete the square. Make sure to add in the right side of the equation whatever was added on the left side of the equation to maintain the equality of both sides. $$(x^2+7x+\color{blue}{(\frac{7}{2})^2})+(y^2+3y+\color{red}{(\frac{3}{2})^2})=-1+\color{blue}{(\frac{7}{2})^2}+\color{red}{(\frac{3}{2})^2} \\(x^2+7x+\frac{49}{4})+(y^2+3y+\frac{9}{4})=-\frac{4}{4}+\frac{49}{4}+\frac{9}{4} \\(x+\frac{7}{2})^2+(y+\frac{3}{2})^2=\frac{50}{4} \\(x+\frac{7}{2})^2+(y+\frac{3}{2})^2=\left(\sqrt{\frac{50}{4}}\right)^2 \\(x+\frac{7}{2})^2+(y+\frac{3}{2})^2=\left(\sqrt{\frac{25(2)}{4}}\right)^2 \\(x+\frac{7}{2})^2+(y+\frac{3}{2})^2=\left(\frac{5\sqrt{2}}{2}\right)^2$$ Therefore, the circle has: center at $(3, 5)$ radius = $\frac{5\sqrt2}{2}$ units

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