For a point P on the standard ellipse $x^2/a^2+y^2/b^2=1 (a>b)$, a triangle $\Delta$PSS' can be formed, it is often found that the minimum cir-cum radius of this triangle is $ae=R_{min}$. Here we wish to point out that this happens only if $1>e^2>1/2$. Otherwise, if $0<e^2<1/2$, the point P gets fixed as $(0,b)$ and $R_{min}= \frac{a^2}{2b}.$
If P$(a \cos t, b \sin t)$, the sides of the triangle PSS' are: PS=$a+ae\cos t$, PS'=$a-ae\cos t$ and SS'$=2ae$ and the area of the triangle is $\Delta= abe \sin t$. Using the formula $R=\frac{abc}{4\Delta}$, for our triangle we can write $$R=\frac{a^2(1-e^2\cos^2 t)}{2b\sin t}= \frac{a^2}{2b} f(t).$$ Next, we notice that $$f'(t)=\frac{\cos t}{\sin^2 t}(e^2\sin^2 t+e^2-1)=0 \implies t_1=\pi/2, \quad t_2=\sin^{-1} \frac{\sqrt{1-e^2}}{e}.$$ Further, $$f''(t)=\csc^3t[(e^2-1) \sin^2 t+2(1-e^2)-e^2\sin^4 t]$$
Case 1: $0<e^2<1/2$: $$f(t_1)=1,f'(t_1)=0 ~\&~ f''(t_1)=1-2e^2 >0 \implies R_{min}=\frac{a^2}{2b}.$$ In this case the equation of the circumcircle passing through $(0,b), (-ae,0), (ae,0)$ is $$x^2+(y-b+r)^2=(\frac{a^2}{2b})^2$$.
Case 2: $1>e^2>1/2$: $\cos t_2= \sqrt{\frac{1-e^2}{2e^2-1}}, f(t_2)=2e\sqrt{1-e^2}, f'(t_2)=0, f''(t_2)= \frac{2e(2e^2-1)}{\sqrt{1-e^2}}>0\implies R_{min}=ae.$ This circle passes through the points $(a \cos t_2, b \sin t_2), (-ae,0), (ae,0).$
To re-emphasize, the case 1 is often overlooked. No question is being asked here. However, you may comment or explain this in some alternate way.