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x^2+y^2-8x-6y+9=0
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When the non-linear terms are of the form \[x^2+y^2+...\] the equation represents an ellipse. By completing the square, we can put it in the standard form: \[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\] The equation becomes \[\frac{(x-4)^2}{(\sqrt{26})^2}+\frac{(y-3)^2}{(\sqrt{26})^2}=1\]
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