I have no idea how I got this circle wedged in this parabola!

At point $(x,y)$ the slope of the parabola is $2x$ (differentiate $y= x^2$ ).

The radius of the circle is perpendicular to the parabola at that point, so it has slope $-1/2x$ .

The slope of the radius is $-d/x$ so $d = 1/2$ .

Then $x^2 + d^2 = R^2$ , so $y = x^2$ implies $y = R^2 -0.25$ and the y-coordinate of the center of the circle is $R^2 + 0.25$ .

Plugging in $R = 3$ gives the answer $9.25$ .

The radius and location of the circle are determined by the tangency with the parabola, so we should find the intersections of their graphs, $y = x^2$ and $(y-a)^2 + x^2 = r^2$ where $a$ is the y-coordinate of the center of the circle, and $r$ is the radius of the circle.

$(x^2-a)^2 + x^2 = r^2$ $x^4-2ax^2+a^2 + x^2 = r^2$ $x^4-(2a-1)x^2+a^2 - r^2 = 0$

Using the quadratic formula to find $x^2$ ,

$x^2 = \frac 1 2 \left ((2a-1) \pm \sqrt{(2a-1)^2 - 4(a^2-r^2)}\right)$

$x^2 = a- \frac 1 2 \pm \sqrt{r^2 - a + \frac 1 4}$

$x = \pm \sqrt{a- \frac 1 2 \pm \sqrt{r^2 - a + \frac 1 4}}$

By inspection, the only way for there to be two and only two different roots is if the determinant of the original quadratic formula is zero.

$r^2 - a + \frac 1 4 = 0$

$a = r^2 + \frac 1 4$

It is simple to verify that these two intersections are indeed points of tangency. This last equation gives us the y-coordinate of the center of the circle as a function of radius. If the area of the circle is $9 \pi$ , then the radius must be $3$ . This gives us a y-coordinate of

$a = 9.25$