Asked by Jen

Graph the Circle
x^2+y^2+4x-2y-20=0

I get:
(x^2+4x)+(y^2-2y)=20
(x^2+4x+4)+(y^2-2y+1)=20+16+4
(x+2)^+(y-1)^2=(40/2)^2

But I don't think its right can you help me with what I did wrong? Thanks

Answers

Answered by helper
(x^2+4x+4)+(y^2-2y+1)=20+16+4

Should be,
(x^2 +4x + 4)+(y^2 - 2y +1 )= 20 + 4 + 1
(x^2 +4x + 4)+(y^2 - 2y +1 )= 25
(x + 2)^2 + (y - 1)^2 = 25




Answered by Jen
Thanks, I see what I did, I squared when adding to the right side
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