What is the radius of a circle given by equation x^2 + y^2 - 2x + 8y - 47 = 0

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asked Oct 3, 2017 121k views

2 Answers

Arnold Schrijver · Answer 1 · 2017-10-10T01:07:05+0000

x^2 + y^2 - 2x + 8y - 47 = 0
x^2 + y^2 - 2x + 8y = 47
(x^2 - 2x) + (y^2 + 8y) = 47
(x^2 - 2(1)x) + (y^2 + 2(4)y) = 47
(x^2 - 2(1)x + 1^2) + (y^2 + 2(4)y + 4^2) = 47 + 1^2 + 4^2
(x - 1)^2 + (y + 4)^2 = 64 = 8^2
r=8

What is the radius of a circle given by equation x^2 + y^2 - 2x + 8y - 47 = 0

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