# Circle in parabola

1. Mar 24, 2007

### Weave

1. The problem statement, all variables and given/known data
A circle with a radius of 1inscribed in the parabola y=x^2, find the center of the circle. The figure shows the circle on the y-axis.

2. Relevant equations
$$y=x^2$$
$$r^2=(x-h)^2+(y-k)^2$$

3. The attempt at a solution
h=0. $$\frac{x^2+(y-k)^2}{r^2}=x^2$$
R^2=1 and the x^2 can be subtracted out leaving:$$(y-k)^2=0$$ take the derivitive of each side, find y, substitute that back in and find k but I end up with other varibles to solve.

Last edited: Mar 24, 2007
2. Mar 24, 2007

### d_leet

How did you get this equation?

3. Mar 24, 2007

### Weave

setting one equation equal to another

4. Mar 24, 2007

### d_leet

And how did you do that? Can you show some work because I cannot see what you set equal to what.

5. Mar 24, 2007

### HallsofIvy

Staff Emeritus
The equation of the circle and the equation of the parabola are describing different curves. They do NOT always have the same (x,y) values!

Unfortunately, we can't see the figure you talk about. In what sense is the circle "inscribed" in the parabola? Is it tangent to the parabola at two points?

From symmetry, it should be obvious that the circle has center on the y-axis. That is, that the equation of the circle is x2+ (y- h)2= r2.

The derivative of y= x2 is y'= 2x and the derivative of y in x2+ (y-h)2= r2 is given by 2x+ 2(y- h)2y'= 0.

You have 4 equations: y= x2, x2+ (y-h)2= r2, y'= 2x, and 2x+ 2(y-h)2= 0 to solve for four unknown values: the (x,y) values of points of intersection, h, and r.