How do you find the center and radius for # x^2 + y^2 - 8x + 14y + 40 = 0#?

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Answer 1 · 2016-06-30T21:01:06

circle centre (4, -7), radius 5

Simple answer: by using algebra and completing the square - TWICE, once for x and once for y.

We're starting with:
#x^2 + y^2 - 8x + 14y + 40 = 0#

So we complete the squares. Will do separately so it's clearer, hopefully

For the x's
#x^2 - 8x = (x-4)^2 - 16#

For the y's
#y^2 + 14y = (y+7)^2 - 49#

pop these back into the original, and we get

#(x-4)^2 - 16 + (y+7)^2 - 49 + 40 = 0#

#(x-4)^2 + (y+7)^2 = 25 color{blue}{= 5^2}#

that's a circle centre (4, -7), radius 5