Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 2 - Graphs and Functions - Quiz - Page 233: 5


center: $(2, -4)$ radius = $\sqrt{17}$

Work Step by Step

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 = $\sqrt{17}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.