a) From Cardano follows $a> \frac{3\sqrt[3]{2}}{2}$ Let $x^2-a=\frac{1}{x}$ has 3 different solutions $k,m,n$ then $k+m+n=0,kmn=1,km+mn+nk=-a$ Line through $(k,\frac{1}{k}),(m, \frac{1}{m})$ : $\frac{y-\frac{1}{k}}{\frac{1}{m}-\frac{1}{k}}=\frac{x-k}{m-k} \to y=-\frac{1}{km}x+\frac{1}{k}+\frac{1}{m}$ Mid perpendicular to $(K,M)$ is $y=kmx-\frac{(k+m)km}{2}+\frac{1}{2k}+\frac{1}{2m}$ x-coordinate of circumcenter is solution of $kmx-\frac{(k+m)km}{2}+\frac{1}{2k}+\frac{1}{2m}=knx-\frac{(k+n)kn}{2}+\frac{1}{2k}+\frac{1}{2n}$ $k(m-n)x+\frac{n-m}{2nm}-\frac{k(m-n)(k+m+n)}{2}=0$ $x=\frac{1}{2mnk}+\frac{(m+n+k)}{2}$ But $m+n+k=0, kmn=1$ so $ x=\frac{1}{2}$ y-coordinate of circumcenter is $y=\frac{km}{2}-\frac{(k+m)km}{2}+\frac{1}{2k}+\frac{1}{2m}=\frac{km+mn+kn}{2}-\frac{-nkm}{2}=\frac{1-a}{2}$ So locus of centers is half-line $x=\frac{1}{2},y< \frac{2-3\sqrt[3]{2}}{4}$