jmc

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, $$t^2=A{O}^2-A{T}^2=A{O}^2-r^2.$$

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, $$A{O}^2={\left(\frac{p+q}{2}\right)}^2+s^2.$$Therefore, $$\begin{array}{rl}{t}^2& =A{O}^2-r^2\\ & ={\left(\frac{p+q}{2}\right)}^2+s^2-{\left(\frac{p-q}{2}\right)}^2-s^2\\ & =pq.\end{array}$$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 $$t^2=O{T}^2=OP\cdot OQ=pq.$$