Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 10 - Analytic Geometry - 10.3 The Ellipse - 10.3 Assess Your Understanding - Page 657: 25


Circle Center: $(0,0)$ r=4 See graph

Work Step by Step

We are given the equation: $x^2+y^2=16$ Put the equation in standard form: $\dfrac{x^2}{16}+\dfrac{y^2}{16}=1$ The equation is in the form: $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$ Determine $h,k,a,b,c$: $h=0$ $k=0$ $a^2=16\Rightarrow a=\sqrt{16}=4$ $b^2=16\Rightarrow b=\sqrt {16}=4$ $c^2=a^2-b^2=16-16=0$ As $a=b$, the equation represents a circle of radius $a=b=4$. The center is: $(h,k)=(0,0)$ Graph the circle:
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