1. **State the problem:** We want to find the value of $c$ such that the parabola $y = -x^2 + 9x - 100$ intersects the line $y = c$ at exactly one point. 2. **Understand the condition for one intersection:** The parabola and the line intersect where their $y$-values are equal: $$-x^2 + 9x - 100 = c$$ Rearranged: $$-x^2 + 9x - (100 + c) = 0$$ 3. **Use the discriminant to find one solution:** For a quadratic equation $ax^2 + bx + d = 0$, the discriminant is $$\Delta = b^2 - 4ad$$ There is exactly one solution if and only if $\Delta = 0$. 4. **Identify coefficients:** Here, $a = -1$, $b = 9$, and $d = -(100 + c)$. 5. **Set discriminant to zero:** $$0 = 9^2 - 4(-1)(-(100 + c))$$ $$0 = 81 - 4(1)(100 + c)$$ $$0 = 81 - 4(100 + c)$$ 6. **Solve for $c$:** $$4(100 + c) = 81$$ $$400 + 4c = 81$$ $$4c = 81 - 400$$ $$4c = -319$$ $$c = \frac{-319}{4}$$ 7. **Final answer:** The value of $c$ is $\boxed{-\frac{319}{4}}$. This corresponds to option C.