$A(a,b)$ and $B(b,-a)$ are two fixed points. If $P(x,y)$ is a moving point such that $$|AP|^2 + |PB|^2 = |AB|^2 \tag1$$ prove that $x^2 + y^2 =(b-a)(x+y)$.
So far I tried to use distance formula in the equation (1). But at the end I could not get the desired outcome. And I think I am not wrong, as it seems to me that it is the only way to solve the problem. Also, could anybody explain to me analytically, when the question says $P(x,y)$ is a moving point.
for P satisfies the condition: $$(x-a)^2+(y-b)^2+(x-b)^2+(y+a)^2=(a-b)^2+(a+b)^2$$ Solve: $$(x^2+a^2-2ax)+(y^2+b^2-2bx)+(x^2+b^2-2bx)+(y^2+a^2+2ay)=2a^2+2b^2$$ $$- a x - 2 b x + x^2 + a y + y^2 = 0$$ $$\left(x-\frac{a+2b}2\right)^2+(y+a/2)^2=\left(\frac{\sqrt{2a^2+4b^2+4ab}}{2}\right)^2$$ The last equation is the locus of point P which apparently is a circle.