jmc

Expanding, we get $$b^2+x^2=a^2-2ax+x^2+b^2-2by+y^2.$$Hence, $$a^2+y^2=2ax+2by.$$Note that $$2by>2y^2\ge y^2,$$so $2by-y^2\ge 0.$ Since $2by-y^2=a^2-2ax,$ $a^2-2ax\ge 0,$ or $$a^2\ge 2ax.$$Since $a>0,$ $a\ge 2x,$ so $$x\le \frac{a}{2}.$$Now, $$a^2\le a^2+y^2=b^2+x^2\le b^2+\frac{a^2}{4},$$so $$\frac{3}{4}{a}^2\le b^2.$$Hence, $${\left(\frac{a}{b}\right)}^2\le \frac{4}{3}.$$Equality occurs when $a=1,$ $b=\frac{\sqrt{3}}{2},$ $x=\frac{1}{2},$ and $y=0,$ so ${\rho }^2=\boxed{\frac{4}{3}}.$

Geometrically, the given conditions state that the points $(0,0),$ $(a,y),$ and $(x,b)$ form an equilateral triangle in the first quadrant. Accordingly, can you find a geometric solution?