The parabola $y = ax^2 + bx + c$ crosses the $x$-axis at $(p,0)$ and $(q,0),$ both to the right of the origin.  A circle also passes through these two points.  Let $t$ be the length of the tangent from the origin to the circle.  Express $t^2$ in terms of one or more of the coefficients $a,$ $b,$ and $c.$

[asy]
unitsize(3 cm);

pair A, O, T;

real func (real x) {
  return ((x - 1)*(x - 2));
}

A = (1.5,-0.4);
O = (0,0);
T = intersectionpoint(Circle(A,abs(A - (1,0))),arc(A/2,abs(A)/2,0,90));

draw(graph(func,0.5,2.5));
draw((-0.5,0)--(2.5,0));
draw((0,-1)--(0,1));
draw(Circle(A,abs(A - (1,0))));
draw(O--T);

label("$t$", T/3, N);

dot(T);
[/asy]
Let $A$ be the center of the circle, let $r$ be the radius of the circle, let $O$ be the origin, and let $T$ be the point of tangency.  Then $\angle OTA = 90^\circ,$ so by the Pythagorean Theorem,
\[t^2 = AO^2 - AT^2 = AO^2 - r^2.\][asy]
unitsize(3 cm);

pair A, O, T;

real func (real x) {
  return ((x - 1)*(x - 2));
}

A = (1.5,-0.4);
O = (0,0);
T = intersectionpoint(Circle(A,abs(A - (1,0))),arc(A/2,abs(A)/2,0,90));

draw(graph(func,0.5,2.5));
draw((-0.5,0)--(2.5,0));
draw((0,-1)--(0,1));
draw(Circle(A,abs(A - (1,0))));
draw(A--T--O--cycle);
draw(rightanglemark(O,T,A,3));

label("$O$", O, NW);
label("$t$", T/3, N);

dot("$A$", A, S);
dot("$T$", T, N);
[/asy]

The center of the circle is equidistant to both $(p,0)$ and $(q,0)$ (since they are both points on the circle), so the $x$-coordinate of $A$ is $\frac{p + q}{2}.$  Let
\[A = \left( \frac{p + q}{2}, s \right).\]Then using the distance from $A$ to $(q,0),$
\[r^2 = \left( \frac{p - q}{2} \right)^2 + s^2.\]Also,
\[AO^2 = \left( \frac{p + q}{2} \right)^2 + s^2.\]Therefore,
\begin{align*}
t^2 &= AO^2 - r^2 \\
&= \left( \frac{p + q}{2} \right)^2 + s^2 - \left( \frac{p - q}{2} \right)^2 - s^2 \\
&= pq.
\end{align*}By Vieta's formulas, $pq = \frac{c}{a},$ so
\[t^2 = pq = \boxed{\frac{c}{a}}.\]Alternatively, by power of a point, if $P = (p,0)$ and $Q = (q,0),$ then
\[t^2 = OT^2 = OP \cdot OQ = pq.\]