College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 6 - Section 6.4 - Summary of the Conic Sections - 6.4 Exercises - Page 619: 15



Work Step by Step

We can see, that both x and y are on the power of two, with the same coefficients. Therefore the equation will be a circle. The given equation is equivalent to: $(x+3)^2+(y-2)^2=16$ Which can be easily transformed into the standard form of the equation of a circle with the centre of $(P;Q)$ and with a radius of $r$: $(x-P)^2+(y-Q)^2=r^2$ Here, it is: $(x-(-3))^2+(y-2)^2=4^2$. Therefore the equation will be a circle with the center of $(-3;2)$ and with a radius of 4.
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