4. Find the equation of the circle that is tangent to the parabola $y=4 x^{2}$ at point $P(1,4)$ and passes through the point $(3,0)$.
4. Since point $P(1,4)$ lies on the parabola, by the general properties of conic sections, the equation of the tangent line to the parabola at point $P$ is $\frac{y+4}{2}=4 x$, which simplifies to $8 x-y-4=0$.
Considering point $(1,4)$ as the point circle $\Gamma_{0}:(x-1)^{2}+(y-4)^{2}=0$, let the equation of the required circle be $(x-1)^{2}+(y-4)^{2}+\lambda(8 x-y-4)=0$.
This circle passes through the point $(3,0)$, thus we find $\lambda=-1$. Therefore, we have $x^{2}+y^{2}-10 x-7 y+21=0$.
Combining this equation with $y=4 x^{2}$ and eliminating $y$, we get
$16 x^{4}-27 x^{2}-10 x+21=0$, which can be factored as $(x-1)^{2}\left(16 x^{2}+32 x+21\right)=0$.
It is easy to see that this equation has only one real root, a double root at $x=1$, thus we conclude that the parabola $y=4 x^{2}$ and the circle $x^{2}+y^{2}-10 x-7 y+21=0$ have only one common point $(1,4)$, meaning they are tangent at point $(1,4)$. Therefore, $x^{2}+y^{2}-10 x-7 y+21=0$ is the equation of the required circle.