19. Let $a, b, c, d$ be real numbers such that
$$
\left\{\begin{array}{l}
a^{2}+b^{2}+2 a-4 b+4=0 \\
c^{2}+d^{2}-4 c+4 d+4=0 .
\end{array}\right.
$$

Let $m$ and $M$ be the minimum and the maximum values of $(a-c)^{2}+(b-d)^{2}$, respectively. What is $m \times M$ ?
19. Answer: 16 .
Complete the squares: $(a+1)^{2}+(b-2)^{2}=1$ and $(c-2)^{2}+(d+2)^{2}=2^{2}$. Each of them represents a circle.
The distance between the two centres is $\sqrt{(2-(-1))^{2}+(-2-2)^{2}}=5$.
So $m=5-1-2=2$ and $M=5+1+2=8$. Thus, $m \times M=16$.