5. The equation of the largest circle above the parabola $y=a x^{2}(a>0)$, which is tangent to the parabola at its vertex, is $\qquad$
5. $x^{2}+\left(y-\frac{1}{2 a}\right)^{2}=\frac{1}{4 a^{2}}$.

Let the required circle be $x^{2}+(y-r)^{2}=r^{2}(r>0)$, and let $P(x, y)$ be any point on the parabola $y=a x^{2}$. Then
$$
\begin{array}{l}
\sqrt{x^{2}+(y-r)^{2}} \geqslant r \text { holds for all } y \geqslant 0 \\
\Leftrightarrow y^{2}+\left(\frac{1}{a}-2 r\right) y \geqslant 0 \text { holds for all } [0,+\infty) \\
\Leftrightarrow r \leqslant \frac{1}{2 a} .
\end{array}
$$

Therefore, the largest circle that satisfies the condition is $x^{2}+\left(y-\frac{1}{2 a}\right)^{2}=\frac{1}{4 a^{2}}$.