Question 783437
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The equation of a circle with center at *[tex \Large \left(x,\,k\right)] and radius *[tex \Large r] is *[tex \Large \left(x\ -\ h\right)^2\ +\ \left(y\ -\ k\right)^2\ =\ r^2].  So to address your situation, you need to make *[tex \Large 2x^2\ +\ 2y^2\ +\ 4y\ =\ 0] look like *[tex \Large \left(x\ -\ h\right)^2\ +\ \left(y\ -\ k\right)^2\ =\ r^2].  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 *[tex \Large x] is trivial since there is no first degree *[tex \Large x] term.  We know from that fact that the value of *[tex \Large h], the *[tex \Large x]-coordinate of the center of the circle, is *[tex \Large 0].


The first step in the process is to divide by the lead coefficient -- the one on the *[tex \Large x^2] term.  Of course, if this is to be a circle, the coefficient on the *[tex \Large y^2] term has to be the same.  So you end up with:


*[tex \LARGE x^2\ +\ y^2\ +\ 2y\ =\ 0]


Next, divide the coefficient on the 1st degree *[tex \Large y] 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:


*[tex \LARGE x^2\ +\ y^2\ +\ 2y\ +\ 1\ =\ 1]


Now the *[tex \LARGE x] term is a perfect square and the *[tex \LARGE y] terms and the constant in the LHS form a perfect square, so factor:


*[tex \LARGE x^2\ +\ \left(y\ +\ 1\right)^2\ =\ 1]


Next, re-write everything so that it fits the circle equation pattern:


*[tex \LARGE \left(x\ -\ 0\right)^2\ +\ \left(y\ -\ (-1)\right)^2\ =\ 1^2]


Now you can determine the center and radius by inspection:


Center:  *[tex \LARGE \left(0,\,-1\right)]


Radius:  *[tex \LARGE 1]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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