A unit circle inscribed in parabola $y=x^2$ has center $(0,5/4)$. Generalize this result to $y=ax^2+bx+c$.

Question

A circle with radius $1$ inscribed in the parabola $y = x^2$. If the coordinates of the center of the circle in this case is found to be $(0, 5/4)$. Can you generalize this problem for an arbitrary quadratic $y = ax^2 + bx + c$? If so, propose a mathematically precise generalization and prove it.

(Recall that the equation of a circle or radius $r$ centered at the point $(h,k)$ is given by $(x −h)^2 +(y −k)^2 = r^2$.)

Answer 1

Rewrite the equation of the general parabola equation as,

$$y = a\left(x+\frac{b}{2a}\right)^2- \frac{b^2}{4a}+c$$

and let the corresponding equation of the circle

$$\left(x+\frac{b}{2a}\right)^2+(y-m)^2=1$$

The center of the circle is $(-\frac{b}{2a}, m)$ with $m$ to be solved next. Combine the two equations above to obtain the quadratic equation for $y$,

$$y^2+\left(\frac 1a -2m\right)y + \frac 1a \left(\frac {b^2}{4a} -c\right)+ m^2 -1=0$$

Since the parabola and the circle are tangential to each other, the discriminant of above quadratic equation is zero, which leads to the solution for the general case,

$$m= \frac{1}{4a}(1+4a^2-b^2+4ac)$$

In the special case of $y=x^2$, the center becomes $(-\frac{b}{2a}, m)=(0,\frac 54)$.