# Circle in parabola

1. Aug 6, 2006

### suspenc3

Hi I am stuck on the following:

Find the centre of a circle with a radius of 1 inscribed in the parabola $$y=x^2$$.

I am kinda stuck:grumpy:

http://img115.imageshack.us/my.php?image=csacacfu8.png
the white line is length=1

Last edited: Aug 6, 2006
2. Aug 6, 2006

### StatusX

I'll do a much simpler example and see if it helps you. Imagine you want to find the point where a unit circle centered at some point on the y-axis is tangent to the x-axis. Clearly we want the center at (0,1) or (0,-1). But another way to find this is to look at where a unit circle centered at (0,h) intersects the line y=0. This is found as the set of simultaneous solutions to:

$$(y-h)^2+x^2=1$$
$$y=0$$

which is just the set of solutions to:

$$x^2+h^2=0$$

If |h|>1, there are no solutions (which reflects the fact that the circle doesn't touch the x-axis), if |h|<1, there are two solutions (the circle straddles rhe x-axis and intersects it in two points), and if h=1 or h=-1, there is one solution, where the circle is tangent, which is just what we expected. Can you figure out how to extend this to your question?

3. Aug 6, 2006

### suspenc3

errr..im still confused, although I do understand what youve done I dont know how to apply it.