 Solution

The Mohr circle has center $C=({\tau }_{xx}+{\tau }_{yy})/2$ on the normal-stress axis and radius $R=\sqrt{({\tau }_{xx}-{\tau }_{yy}{)}^2/4+{\tau }_{xy}^2}$. In the compressive regime $C<0$, yield occurs when the circle first touches either friction line $F=\pm (-N)\tan \varphi$. The perpendicular distance of its center from either line is $|C|\sin \varphi$. Tangency therefore gives $R^2=C^2{\sin }^2\varphi$, or $C^2-R^2=C^2{\cos }^2\varphi$. But $C^2-R^2={\tau }_{xx}{\tau }_{yy}-{\tau }_{xy}^2$. ThusThe compressive sign restriction remains part of this equivalence: squaring alone would also produce algebraic branches that do not represent cohesionless compressive friction.