SOLUTION: 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 wha

Question 403004: 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
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
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+4+1
(x+2)^2 + (y-1) = 25
-----
(x+2)^+(y-1)^2 = 5^2
------
Center at (-2,1)
Radius = 5
=====================
Cheers,
Stan H.
===========
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

You are right -- that is not right. Why did you add 16 and 4 to the RHS when you only added 4 and 1 to the LHS? There is no factor of 4 in front of either set of parentheses in the LHS. Furthermore, if 40 was the correct value in the RHS, then the RHS would be properly expressed as

Center at , radius

John

My calculator said it, I believe it, that settles it

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