Find the center and the radius of the circle x^2 + y^2 - 2ax + 4by = 0.
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Does every equation of the form x^2 + mx + y^2 + ny = p represent a circle? Explain.
Complete the square on x and y and gather terms into something like:
(x+p)^2 + (y+q)^2 = r^2
Then (-p, -q) is the centre and r is the radius.
It's not a circle if r^2 = 0 in which case it's a single point.
Or when r^2 is negative in which case there are no points at all.
Let $\displaystyle A,B,C\in\mathbb{R}$,
then
$\displaystyle x^{2}+y^{2}+Ax+By+C=0$
represents a circle provided that
$\displaystyle A^{2}+B^{2}-4C>0$
and the circle is centered at
$\displaystyle \bigg(-\frac{A}{2},-\frac{B}{2}\bigg)$
and the radius is
$\displaystyle \frac{\sqrt{A^{2}+B^{2}-4C}}{2}.$
Also you may try squaring the given equation to get $\displaystyle (x-m)^{2}+(y-n)^{2}-r^{2}=0$, which is centered at $\displaystyle (m,n)$ with a radius of $\displaystyle r$.