Answer
The center of the circle is at the point $(0, \frac{5}{4})$
Work Step by Step
Let $x_0$ be the positive x-coordinate where the circle intersects the parabola.
Then the point $(x_0, x_0^2)$ is a point of intersection.
The slope of the parabola at this point is $2 x_0$
Since the circle is tangent to the parabola at this point, then the line connecting the point $(x_0,x_0^2)$ to the center of the circle is $-\frac{1}{2x_0}$
Let $b$ be the vertical distance from the point $(x_0, x_0^2)$ and the center of the circle. Then $b = \sqrt{1-x_0^2}$
We can find $x_0$:
$\frac{b}{-x_0} = -\frac{1}{2x_0}$
$b = \frac{1}{2}$
$\sqrt{1-x_0^2} = \frac{1}{2}$
$1-x_0^2 = \frac{1}{4}$
$x_0^2 = \frac{3}{4}$
$x_0 = \frac{\sqrt{3}}{2}$
The center of the circle is at the point $(0, x_0^2+b)$
We can find $x_0^2+b$:
$x_0^2+b = x_0^2+\sqrt{1-x_0^2}$
$x_0^2+b = (\frac{\sqrt{3}}{2})^2+\sqrt{1-(\frac{\sqrt{3}}{2})^2}$
$x_0^2+b = \frac{3}{4}+\sqrt{\frac{1}{4}}$
$x_0^2+b = \frac{3}{4}+\frac{1}{2}$
$x_0^2+b = \frac{5}{4}$
The center of the circle is at the point $(0, \frac{5}{4})$