4. Given the parabola $y=a x^{2}+b x+c(a \neq 0)$ intersects the $x$-axis at points $P, Q$ which are on opposite sides of the $y$-axis, and the circle with diameter $P Q$ intersects the $y$-axis at $M_{1}(0,4)$ and $M_{2}(0,-4)$. If the vertex of the parabola is $\left(-\frac{b}{2 a},-\frac{1}{4 a}\right)$, then the curve on which the point $(b, c)$ lies is
(A) Circle
(B) Ellipse
(C) Hyperbola
(D) Parabola
4. B.

As shown in Figure 3, let the x-coordinates of the points where the parabola intersects the x-axis be $x_{1}$ and $x_{2}$. Then, $\frac{c}{a}=x_{1} x_{2}<0$. In the right triangle $\triangle P M_{i} Q (i=1,2)$, $M_{i} O$ is the altitude to the hypotenuse, so we have
$$
|O P| \cdot|O Q|=\left|O M_{i}\right|^{2},
$$

which means $\left|x_{1} x_{2}\right|=16$.
Thus, $\frac{c}{a}=-16$.
From the y-coordinate of the vertex of the parabola, we get $\frac{4 a c-b^{2}}{4 a}=-\frac{1}{4 a}$. Therefore, $b^{2}-4 a c=1$.
Substituting equation (1) into the above equation to eliminate $a$ yields $b^{2}+\frac{c^{2}}{4}=1$.