28. There is a point $P(x, y)$ on the plane. If the distance from point $P$ to $(3,0)$ and the distance from point $P$ to $(0,3)$ sum up to 6, then point $P$ must satisfy ( ).
(A) $3(x-2)^{2}+3(y-2)^{2}+3 x y=24$
(B) $3(x-2)^{2}+3(y-2)^{2}+2 x=24$
(C) $3(x-2)^{2}+3(y-2)^{2}+3 x y=18$
(D) $3(x-2)^{2}+3(y-2)^{2}+2 x y=18$
(E) *
28. (B).

From the given, we have
$$
\begin{array}{l} 
\sqrt{(x-3)^{2}+y^{2}}-\sqrt{x^{2}+(y-3)^{2}}=6 \\
\Rightarrow 2 \sqrt{x^{2}+(y-3)^{2}}=6+(x-y) \\
\Rightarrow 3\left(x^{2}+y^{2}\right)-12(x+y)+2 x y=0 \\
\Rightarrow 3(x-2)^{2}+3(y-2)^{2}+2 x y=24 .
\end{array}
$$