The maximum value of  $P=(1-x_1)(1-y_1)+(1-x_2)(1-y_2)$ 

where  ${x_1}^2+{x_2}^2={y_1}^2+{y_2}^2=2$

let  $x_1=\sqrt{2}\cos \theta$
$y_1=\sqrt{2}\sin \theta$
$x_2=\sqrt{2}\sin \theta$
$y_2=\sqrt{2}\cos \theta$
$P=2-x_1-y_1-x_2-y_2+x_1{y}_1+x_2{y}_2$
$=2-\sqrt{2}(\cos \theta +\sin \theta )-\sqrt{2}(\cos \theta +\sin \theta )+2\sin \theta \cos \theta +2\sin \theta \cos \theta$$=2-4\cos (\theta -\frac{\pi }{4})+2\sin 2\theta$
$=2+4+2=8$