Question 783437: the equation is: 2x^2+2y^2+4y=0 and I am supposed to find the center and the radius of the circle as well as the x and y intercepts but I am not sure how to find the center and radius. If you could help me find those I can find the x and y intercepts. Thanks
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 2x^2+2y^2+4y=0 and I am supposed to find the center and the radius of the circle
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Complete the square on the x-terms and on the y-terms:::
2(x^2) + 2(y^2+2y) = 0
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2(x-0)^2 + 2(y^2+2y+1) = 2
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2(x-0)^2 + 2(y+1)^2 = 2
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(x-0)^2 + (y+1)^2 = 1
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Center: (0,-1)
Radus : sqrt(2)
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Cheers,
Stan H.
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Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
The equation of a circle with center at ) and radius  is ^2\ +\ \left(y\ -\ k\right)^2\ =\ r^2) . So to address your situation, you need to make  look like . Actually, much easier than it looks.
The process requires you to complete the square on each of the two variables. For this particular problem, dealing with  is trivial since there is no first degree term. We know from that fact that the value of  , the -coordinate of the center of the circle, is  .
The first step in the process is to divide by the lead coefficient -- the one on the  term. Of course, if this is to be a circle, the coefficient on the  term has to be the same. So you end up with:
Next, divide the coefficient on the 1st degree  term by 2 and square the result. Add that result to both sides of the equation. 2 divided by 2 is 1. 1 squared is 1, so:
Now the  term is a perfect square and the  terms and the constant in the LHS form a perfect square, so factor:
^2\ =\ 1)
Next, re-write everything so that it fits the circle equation pattern:
^2\ +\ \left(y\ -\ (-1)\right)^2\ =\ 1^2)
Now you can determine the center and radius by inspection:
Center: )
Radius: 
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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