How do you graph #x^2+y^2-6x-14y+42=0#?

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Answer 1 · 2015-09-01T17:58:17

This formula describes a circle, which center is in #O=(3,7)# and radius is #4#

To find this out you have to transform the formula to:

#(x-a)^2+(y-b)^2=r^2#.

In this formula #a# and #b# are coordinates of the center and #r# is the radius

#x^2+y^2-6x-14y+42=0#

#(x^2-6x color(red)(+9)) +(y^2-14y color(red)(+49)) color(red)(-9-49) +42=0#

#(x-3)^2+(y-7)^2-16=0#

#(x-3)^2+(y-7)^2=16#