Already so many answers, but I haven't seen my favorite one posted, so here's another.
The vertex occurs on the vertical line of symmetry, which is not affected by shifting up or down. So subtract $c$ to obtain the parabola $y=ax^2+bx$ having the same axis of symmetry. Factoring $y=x(ax+b)$, we see that the $x$-intercepts of this parabola occur at $x=0$ and $x=-\frac{b}{a}$, and hence the axis of symmetry lies halfway between, at $x=-\frac{b}{2a}$.
Why is it that when we convert radians to degrees we multiply radians $\times \frac{180}π$ , but when we convert slope per radians to slope per degrees we have to multiply the inverse conversion formula slope per radians $\times \frac{π}{180}$
If we want to know an hour in terms of minutes, we multiply 1 hour $\times \frac{60}{1}$, given the result in minutes. If we want to know how convert 180 minutes, we divide $180$ minutes by $60$, i.e., multiply $180 \times \frac 1{60}$.
You'll find this phenomenon in any conversion: To convert temperature in degrees Celsius to temp in Fahrenheit, we have $F = \frac 95 C + 32$. To convert to from F to C, we need to invert this: $C = \frac 59(F-32)$
$R \text{ radians}\;\times \dfrac{180^\circ}{\pi \;\text{radians}} = \dfrac{R\times 180^\circ}{\pi}$.
"Radians" cancels as the unit, leaving a numeric value expressed in degrees.
$D \text{ degrees}\; \times \dfrac{\pi \;\text{radians}}{180\; \text{degrees}} = \dfrac{D\pi\;}{180}\;\text{ radians}$.
"Degrees" cancels as the unit, leaving the value expressed in radians.
Moved from comments:
Note that slope per radian is a ratio: $\;\dfrac{\text{slope}}{\text{radians}}.\;$ So to obtain "slope per degree", you need to have $π$ radians in the numerator, to cancel the unit "radians" from the denominator, and $180$ degrees in the denominator, to end with slope/degrees.
Slope itself is not a "unit" per se, meaning it isn't a degree, or radian, a mm, or foot. It is unit-free: even if we assign distance units (meters, say) to displacement: e.g. $Δ\,y\text{m}=y_2\,\text{m}−y_1\,\text{m}$ and $Δ\,x\,\text{m}=x_2\text{m}−x_1\,\text{m},$ then we have $$\text{slope}\,= \dfrac{Δy\,\text{ m}}{Δx\,\text{m}}=\frac{Δy}{Δx},$$ you see that the units "m: meter" attached to "change in y" and "change in x" cancel in the ratio defining slope, leaving us with a unit-free scalar which slope really is.
Best Answer
You are correct.
What you are doing by substituting $y$ for $x^2$ is solving for the y-values of the points of intersection of the curves $x^2 + (y - c)^2 = r^2$ and $y = x^2$.
If the discriminant is equal to zero, then the parabola and circle are tangent. If the discriminant were positive, the parabola and circle would intersect at four points (since there are two $x$-values for each value of $y$ that solves the quadratic equation $y + (y - c)^2 = r^2$). If the discriminant were negative, the circle and parabola would not intersect.
Since \begin{align*} y + (y - c)^2 & = r^2\\ y + y^2 - 2cy + c^2 - r^2 & = 0\\ y^2 + (1 - 2c)y + c^2 - r^2 & = 0 \end{align*} has discriminant \begin{align*} \Delta & = b^2 - 4ac\\\ & = (1 - 2c)^2 - 4(1)(c^2 - r^2)\\ & = 1 - 4c + 4c^2 - 4c^2 - 4r^2\\ & = 1 - 4c - 4r^2 \end{align*} and the two curves are tangent when $\Delta = 0$, $4c = 1 - 4r^2$.
The reason for your friend's confusion is that for each value of $y > 0$ that satisfies the equation $y + (y - c)^2 = r^2$, there are two points $(\sqrt{y}, y)$ and $(-\sqrt{y}, y)$ where the parabola $y = x^2$ intersects the curve $x^2 + (y - c)^2 = r^2$.