(4) In the Cartesian coordinate system, given points $A(1,2)$ and $B(4,1)$. For a moving point $P(x, y)$ on the circle $x^{2}+y^{2}=25$ forming a triangle with $A, B$, the maximum value of the area of triangle $ABP$ is $\qquad$.
(4) $\frac{1}{2}(7+5 \sqrt{10})$ Hint: Calculate the area of triangle $ABP$ with $AB$ as the base, then the size of the area is determined by the height of the triangle. Note that points $A$ and $B$ are both inside the circle, so the maximum height of triangle $ABP$ with $AB$ as the base is the sum of the distance from the line containing segment $AB$ to the origin and the radius of the circle.

The length of segment $AB$ is $\sqrt{(4-1)^{2}+(1-2)^{2}}=\sqrt{10}$. The equation of the line containing segment $AB$ is $x+3 y-7=0$, and its distance to the origin is $\frac{|-7|}{\sqrt{1^{2}+3^{2}}}=\frac{7}{\sqrt{10}}$. Therefore, the maximum height of the triangle is $\frac{7}{\sqrt{10}}+5$. Hence, the maximum area of triangle $ABP$ is
$$
\frac{1}{2} \times \sqrt{10} \times\left(\frac{7}{\sqrt{10}}+5\right)=\frac{1}{2}(7+5 \sqrt{10}) .
$$