## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Chapter 4 - Chapter Test - Page 355: 17

#### Answer

Vertex: $(4,-6), (-2,6)$ and center: $(1,-6)$ Focus: $(1+\sqrt 6, -6), (1 -\sqrt 6, -6)$

#### Work Step by Step

The standard form of the equation of the circle is: $(x^2-2x+1)+3(y^2+12y+36)=-100+1+108$ or, $(x-1)^2+3(y+6)^2=9$ So, we see that the vertex is: $(h \pm a , k)=(4,-6), (-2,6)$ The center is: $(1,-6)$ $c=\sqrt {a^2-b^2}=\sqrt{6}$ Focus: $(h \pm c , k)=(1+\sqrt 6, -6), (1 -\sqrt 6, -6)$

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