Question

Let $a,b,c,d$ be real numbers such that $\{\begin{array}{c}{a}^2+b^2+2a-4b+4=0\\ c^2+d^2-4c+4d+4=0\end{array}.$ Let $m$ and $M$ be the minimum and the maximum values of $(a-c{)}^2+(b-d{)}^2,$ respectively. The value of $m\times M$ is
(correct answer + 3, wrong answer 0)

Answer 1

$a^2+b^2+2a-4b+4=0$
$\Rightarrow (a+1{)}^2+(b-2{)}^2=1^2$

Similarly, for $c^2+d^2-4c+4d+4=0$
$(c-2{)}^2+(d+2{)}^2=2^2$
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, $m\times M=2\times 8=16$