Answer
$\left(0,\frac{5}{4}\right)$
Work Step by Step
The x-coordinate of the center is $0$ as shown in the sketch so the center of the circle has the form:
$$(0,c)$$
The equation of the circle is:
$$x^{2}+(y-c)^{2}=1$$
Let $(a,a^{2})$ be the the point of tangency of the circle and parabola.
so:
$$a^{2}+(a^{2}-c)^{2}=1$$
The slope of the tangent line to the graph of the parabola is:
$$m=2x|_{x=a}=2a$$
The slope of the tangent line to the circle can be found by the implicit differentiation of the equation of the circle:
$$2x+2(y-c)y'=0$$
$$y'=-\frac{x}{y-c}$$
$$m=y'|_{x=a,y=a^{2}}=-\frac{a}{a^{2}-c}$$
Since both curves share the same tangent line so their slopes are equal:
$$2a=-\frac{a}{a^{2}-c} \to a^{2}-c=-\frac{1}{2} $$
Substitute this value into the equation of the circle at $a$:
$$a^{2}+(a^{2}-c)^{2}=1 \to a^{2}+\frac{1}{4}=1 \to a^{2}=\frac{3}{4} $$
thus:
$$a^{2}-c=-\frac{1}{2} \to \frac{3}{4} -c=-\frac{1}{2} \to c=\frac{5}{4}$$
thus the center of the circle is:
$$\left(0,\frac{5}{4}\right)$$
