Circle tangent to a parabola this is a calculus-geometry question that I found - 
A circle of radius $1$ is tangent to the parabola $y = x^2 -2x + 1$ at two points. Find the coordinates of the center of the circle.
My attempt:
Let the center of the circle be $(a, b)$. Since there are two intersection points, we have:
$(x - a)^2 + (y - b)^2 - 1 = x^2 - 2x + 1$
$x^2 - 2ax + a^2 + y^2 - 2ay + b^2 = x^2 - 2x$
$-2ax + a^2 + y^2 - 2ay + b^2 = -2x$
I am unsure of how to proceed; I think I am approaching this problem wrong because I did not take the derivative. The correct answer is $(1, 5/4).$
 A: It is simpler to translate the parabola by $1$ towards the left,  solving the problem for the parabola $y=x^2$ and then re-translating the final result by one unit towards the right. In this way, we can search a  circle with its center on the $y $-axis, which typically has an equation of the form 
$$x^2 +(y-k)^2=1$$
where $k $ is the $y $-coordinate of the lower point of the circle at its inferior intersection with the $y $-axis. The points of intersection between the parabola  and the circle are the solutions of the system
$$\left\{\begin{array}{lll}
x^2 +(y-k)^2=1 \\
y=x^2
\end{array}\right.$$
The solutions for $y $ are
$$y = \frac {1}{2} \left(2 k \pm \sqrt{5 - 4 k} - 1 \right) $$
Because we are searching for points where the curves are tangent, the two solutions of $y $ have to coincide. This occurs for $k=\frac {5}{4} $. This leads to the circle 
$$x^2 +(y-\frac {5}{4})^2=1$$
Re-translating towards the right we get the final equation of the circle
$$(x-1)^2 +(y-\frac {5}{4})^2=1$$
whose center is in $(1,\frac {5}{4}) $. Here is a graph of the two tangent curves. 
A: The parabola $y = x^2 - 2x +1$ can be written as $y = (x-1)^2$. You should be able to convince yourself that the center of the circle has to lie on the parabola's axis of symmetry, $x=1$.
