Calculus 8th Edition

Published by Cengage

Appendix C - Graphs of Second-Degree Equations - C Exercises - Page A23: 5

Answer

center,$(h,k)=(2,-5)$ and $r=4$

Work Step by Step

An equation of the circle with center $(h,k)$ and radius $r$ is given as: $(x-h)^{2}+(y-k)^{2}=r^{2}$ ...(1) Given: $x^{2}+ y^{2}-4x +10y +13 = 0$ The above equation can be written in the standard equation of the circle as follows: $(x-2)^{2}+ (y-(-5))^{2} = 4^{2}$ Compare it with equation (1), we have Hence, center,$(h,k)=(2,-5)$ and $r=4$

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