# Center and radius of circle question

For a circle that has the equation $(x-a)^2 + (y-b)^2 = r^2$ , I know it has center $(a,b)$ and radius $r$.

But what happens if the equation is $x^2 + (y-b)^2 = r^2$ ? With $b$ not equal to $0$ and $a=0$ ? How would the "circle" look like?

Thanks so much!

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What happens when you put $a=0$ in the first line? – Raskolnikov Oct 21 '11 at 9:14
May be you wish to know "The center is on the y-axis"... – Tapu Oct 21 '11 at 9:19
Your equation can be rewritten as $(x-0)^2+(y-b)^2=r^2$. – André Nicolas Oct 21 '11 at 11:43

if $C(0,b)$ and $r>|b|$
There's nothing special about the coordinate $a=0$. The coordinates of the centre are the result of an arbitrary choice of coordinate system. Thus one coordinate of the centre being zero doesn't make any difference with respect to the form of the circle, and there's no reason to put "circle" in quotation marks -- this is a circle like any other circle, and it looks like any other circle; in fact any other circle can be described with $a=0$ for a suitable choice of the coordinate system.
The situation is quite different with respect to $r$; you would be quite justified in placing "circle" in quotation marks if the question was what happens for $r=0$.