Precalculus (6th Edition)

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

Chapter 2 - Graphs and Functions - 2.2 Circles - 2.2 Exercises: 27

Answer

The graph fo the given equation is a circle with: center at $(-3, -4)$ radius = 4 units

Work Step by Step

RECALL: (1) The center-radius form of a circle's equation is $(x-h)^2+(y-k)^2=r^2$ 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 $(x+b)^2$. Subtract $9$ to both sides of thee equation to obtain: $x^2+y^2+6x+8y=-9$ Group terms with the same variables to obtain: $(x^2+6x)+(y^2+8y)=-9$ 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. $(x^2+6x+\color{red}{(\frac{6}{2})^2}) + (y^2+8y+\color{red}{(\frac{8}{2})^2}) = -9 +\color{red}{(\frac{6}{2})^2+(\frac{8}{2})^2} \\(x^2+6x+9)+(y^2+8y+16)=-9+9+16 \\(x^2+6x+9)+(y^2+8y+16)=16 \\(x+3)^2+(y+4)^2=4^2$ The equation above is in the same form as the formula/equation in (1) above. Thus, its graph is a circle with center at $(-3, -4)$ and a radius of $4$.
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.