Unit circle shifted upwards so it is tangent the graph of $f(x)=x^{2}$

Question

How can we find k so that $x^{2}+(y-k)^{2}$ is tangent to the graph of $f(x)=x^2$?

Answer 1

Substantial hint:

Suppose you knew $k$. Then at the (right hand) point where the graphs touch -- call it $(a,b)$, you'd know three things:

1. $b = a^2$
2. $a^2 + (b-k)^2 = 1$
3. $\frac{b-k}{a} = \frac{-1}{2a}$

The last follows by drawing a line from $(a, b)$ to the circle center at $(0, k)$. That line has slope $\frac{b-k}{a}$. It's also perpendicular to the tangent line to the parabola at $(a, b)$, which has slope $2a$. And the product of the slopes of perpendicular lines is always $-1$.

Now you just have to solve those three simultaneous equations.

Answer 2

The points of intersection of the parabola and the circle are the solutions of the pair of equations $$y=x^2 \text{ and } x^{2}+(y-k)^{2}=1 $$ or, equivalently, $$y=x^2 \text{ and } y+(y-k)^{2}=1 $$ The circle is tangent to the parabola if the quadratic equation for $y$ has exactly one solution.