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
Algebra ->
Circles
-> 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
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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.
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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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