Prove by contradiction and set $r^2+c^2=\left(\frac{p}{q}\right)^2$ with $p,q \in \mathbb{Z}^+$ and $\gcd{(p,q)}=1$. It follows that $r^2 \in \mathbb{Q}$, but $r = \frac{-ar^2-c}{b}$ so $r \in \mathbb{Q}$. Set $r = \frac{m}{n}$ with $m,n \in \mathbb{Z}^+$ and $\gcd{(m,n)}=1$. We now have $\sqrt{\left(\frac{m}{n}\right)^2 + c^2} = \frac{p}{q}$, so $\frac{\sqrt{m^2+c^2n^2}}{n} = \frac{p}{q}$. Since $\gcd{(m^2+c^2n^2,n)} = \gcd{(m^2,n)}=1$, both fractions are irreducible, and $n=q$. Now \[\frac{p}{n} = \sqrt{r^2+c^2} = \sqrt{r^2 + (-ar^2-br)^2} = \frac{m}{n}\sqrt{1+\left(\frac{am}{n}+b\right)^2}\] So $m\sqrt{n^2+(am+bn)^2}=pn \in \mathbb{Z}$. $n^2+(am+bn)^2$ is an integer, so $\sqrt{n^2+(am+bn)^2}$ is either an integer or irrational; it cannot be the latter. Thus \[n|m\sqrt{n^2+(am+bn)^2} \Longrightarrow n|\sqrt{n^2+(am+bn)^2} \Longrightarrow n^2|(am+bn)^2 \Longrightarrow n|a\] Let $ny = a$, $y \in \mathbb{Z}$; then $m\sqrt{1+(my+b)^2}=p$. $\sqrt{1+(my+b)^2}$ must be an integer, but the only two perfect squares that are 1 apart are 0 and 1. Then $m=p$, so $\frac{p}{q}=\frac{m}{n}=r$ and $\sqrt{r^2+c^2}=r$, contradicting the fact that $c \neq 0$.

Assume $\sqrt{r^2+c^2} \in \mathbb{Q}$. When $b=0$, we have $r = \sqrt{-\frac{c}{a}}$, so $a$ and $c$ are of opposite sign. We know \[ \sqrt{r^2+c^2} = \sqrt{c^2-\frac{c}{a}} = \frac{1}{a}\sqrt{a^2c^2-ac} \in \mathbb{Q}\] It follows that $a^2c^2-ac$ must be a perfect square. But $a^2c^2-ac$ is in between $(ac-1)^2$ and $(ac)^2$ with equality holding if either $ac=0$ or $ac=1$. The former cannot be true, as specified in the problem, and the latter cannot be true because $a$ and $c$ are of opposite sign.