Let $A=(1,2),B=(3,4)$ and let $C=(x,y)$ be a point such that $(x-1)(x-3)+(y-2)(y-4)=0$. If $\operatorname{ar}(\Delta ABC)=1,$then what is the maximum number of positions of $C$ in the $x$-$y$ plane?
Solution
$(x-1)(x-3)+(y-2)(y-4)=0\implies AC \perp BC \implies \angle ACB =90^0 \implies$ $C$ is on the circle whose diameter is $AB=2\sqrt 2$
What next??