${x}^(2) + {y}^(2) - 4x + 2y - 31 = 0 \\ {x}^(2) + {y}^(2) - 4x + 2y = 31 \\ ({x}^(2) - 4x + 4) +( {y}^(2) +2y + 1) = 31 + 4 + 1 \\ {(x - 2)}^(2) + {(y + 1)}^(2) = 36 = {6}^(2)$

so from the equation

center ( 2 , -1 )

radius 6

OfLettersAndNumbers answered May 3, 2020

by OfLettersAndNumbers

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1 vote

Answer:

25y-4x

Explanation:

AKIWEB answered May 8, 2020

by AKIWEB

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X^2+y^2-4x+2y-31=0 what are the center and radius of the circle? Show work

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