= 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 $F=\pm(-N)\tan\phi$. The perpendicular distance of its center from either line is $|C|\sin\phi$. Tangency therefore gives $R^2=C^2\sin^2\phi$, or $C^2-R^2=C^2\cos^2\phi$. But $C^2-R^2=\tau_{xx}\tau_{yy}-\tau_{xy}^2$. Thus
$$
\boxed{4(\tau_{xx}\tau_{yy}-\tau_{xy}^2)=(\tau_{xx}+\tau_{yy})^2\cos^2\phi.}
$$
The compressive sign restriction remains part of this equivalence: squaring alone would also produce algebraic branches that do not represent cohesionless compressive <friction>.
\Image[/past-exam-of-the-elasticity-and-plasticity-course-of-the-university-of-oxford/2009-mohr-coulomb-tangency.png]
{title=Mohr-circle tangency to the cohesionless friction envelope in tension-positive coordinates}
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