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How do you find the point for a circle and find the radius when $x^2$ has a co-efficient?

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I'm assuming you mean the point which is the center of the circle? Could you give an example of what you mean? What equation are you working with, e.g.? – amWhy Feb 19 '13 at 14:25
you can divide the whole equation by the coefficient of $x^2$ and then treat it the way you do any of your "normal" equations. – Maesumi Feb 19 '13 at 14:29

For an equation of the form $$ax^2 + by^2 = c^2$$ unless $a = b$, you do not have a circle, but rather, an ellipse.

If $a = b$, then you do have a circle, and you can rewrite your equation as $$x^2 + y^2 = \left(\frac ca\right)^2\tag{1}$$

In this case $(1)$, the center of the circle is the origin, and the radius is $\dfrac ca$.

In general, the equation of a circle with center $(x_0, y_0)$ and radius $r$ is given by:$$(x- x_0)^2 + (y-y_0)^2 = r^2$$

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