# 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 $(1, 5/4).$

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.

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, $x=1$.