Let A(–2, 2) and B(2, –2) be two points and AB subtends an angle of 45° at any points P in the plane  in such a way that area of $\mathrm{△}\mathrm{APB}$is 8 square unit, then number of possible position(s) of P is

Since AB=4 and area of △APB=8

∴,$\therefore \frac{1}{2}\times 4\times \mathrm{h}=8\Rightarrow \mathrm{h}=4$ where h is  the height of △APB

From the figure it is clear that P lies on a circle of radius $2\sqrt{2}$

​ units, with AB as its chord, so there are four possible positions of P