Question 310855
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You know that the equation represents a circle because the coefficients on the squared variables are equal.


To find the center, complete the square.  Since there is no first degree y term, the square is already complete on y, hence the y coordinate of the center is zero.


Divide the equation by the coefficient on x.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ y^2\ -\ \frac{3}{2}x\ =\ 0]


Divide the coefficient on the first degree x term by 2, then square the result.  Add that result to both sides of the equation.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ \frac{3}{2}x\ +\ \frac{9}{16}\ +\ y^2\ =\ \frac{9}{16}]


Factor the perfect square trinomial in x:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ -\ \frac{3}{4}\right)^2\ +\ y^2\ =\ \left(\frac{3}{4}\right)^2]


The center is at *[tex \Large \left(\frac{3}{4},0\right)] and the radius is *[tex \Large \frac{3}{4}]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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