Mathematical Immaturity

14.21 Miscellaneous Review Exercises

• Given that the line $y = c$ intersects the parabola $y = x^2$ at two points. Find the radius of the circle passing through these two points and through the vertex of the parabola. The radius you determine depends on $c.$ What happens to this radius as $c \to 0$?
• What are the $x$-coordinates of the two points at which the line $y = c$ intersects the graph of $y = x^2?$ What does this imply about the center of the circle? What happens to the radius if $c = 0;$ how does this differ from the radius in the limit as $c \to 0?$
• The line $y = c$ intersects the parabola $y = x^2$ at the points $(-\sqrt{c}, c)$ and $(\sqrt{c}, c).$ If these two points are both on the circle with center at $P = (p, q),$ then it follows that \begin{align*} \|P - (-\sqrt{c}, c)\| = \|P - (\sqrt{c}, c)\| \end{align*} Or, in other words, \begin{align*} \left(p + \sqrt{c}\right)^2 + (q - c)^2 &= \left(p - \sqrt{c}\right)^2 + (q - c)^2 \\ \\ \left(p + \sqrt{c}\right)^2 &= \left(p - \sqrt{c}\right)^2 \end{align*} But, if $c \neq 0,$ then this implies $p = 0.$ (Note: If $c = 0,$ then the two points converge at the vertex of the parabola and the circle has radius $0.$) Now, given that the circle also goes through the vertex of the parabola (ie, the origin), this means $\|P\| = \|P - (-\sqrt{c}, c)\|$ and $\|P\| = \|P - (\sqrt{c}, c)\|.$ And since the $x$-coordinate of $P$ is zero, we get: \begin{align*} \\ q^2 &= c + (q - c)^2 \\ \\ &= c + q^2 - 2qc + c^2 \end{align*} Rearranging variables and simplifying, we find that \begin{align*} q &= \frac{1 + c}{2} \end{align*} Which means that the center of the circle $P$ is at $\left(0, \frac{1 + c}{2}\right).$ But since the circle goes through the origin, $\|P\|$ is the radius of the circle. Thus, the radius $r$ of the circle is $\frac{1 + c}{2}.$ And in the limit, as $c \rightarrow 0,$ $r \to \frac{1}{2}.\,\blacksquare$