In polar coordinates, the electric potential of a ring is represented by the next relation $$ \frac{\lambda}{4\pi\varepsilon_0}\frac{2R}{|r-R|}\left( F\left(\pi -\frac{\theta}{2}\Big|-\frac{4 r R}{(r-R)^2} \right) - F\left( -\frac{\theta}{2}\Big|-\frac{4 r R}{(r-R)^2} \right) \right). $$ Which, for a ring, has the same value if $r$ is constant (equipotential lines are circles).
So, an elliptic integral seems to have the next property: $$ F(\pi-x|y)+F(x|y) = k(y), $$ where $k$ is a constant that depends only on $y$.
My question, is there a way to find $k(y)$?
Also, I didn't prove the property, but the symmetry of the problem indicates that the statement is true. It would be interesting if someone could prove that result.
I am writing an answer based on my comments to add some more details.
Let us use the definition $$F(x\mid y) =\int_0^x\frac{dt}{\sqrt{1-y\sin^2t}}\tag{1}$$ And then consider the sum $$f(x) =F(x\mid y) +F(\pi - x\mid y) \tag{2}$$ We have $$F(\pi-x\mid y) =\int_0^{\pi - x} \frac{dt} {\sqrt{1-y\sin^2t}}$$ which can be transformed into $$\int_{\pi} ^{x} \frac{-du}{\sqrt{1-y\sin^2u}}=\int_x^{\pi}\frac{dt}{\sqrt{1-y\sin^2t}}$$ via the substitution $u=\pi - t$. And then we have $$f(x) =\int_0^x\frac{dt}{\sqrt{1-y\sin^2t}}+\int_x^{\pi}\frac{dt}{\sqrt{1-y\sin^2t}}=\int_{0}^{\pi}\frac{dt}{\sqrt {1-y\sin^2t}}$$ and the above equals $$2\int_0^{\pi/2}\frac{dt}{\sqrt{1-y\sin^2t}}=2K(y)$$ because the integrand does not change when $t$ is replaced by $\pi-t$.
Another approach is to differentiate the expression $f(x) $. We have $$\frac{d} {dx} F(x\mid y) =\frac{1}{\sqrt{1-y\sin^2x} }$$ and $$\frac{d} {dx} F(\pi - x\mid y) =-\frac{1}{\sqrt{1-y\sin^2(\pi-x)}}=-\frac {1}{\sqrt{1-y\sin^2x}}$$ so that $f'(x) =0$ and hence $f(x) $ is independent of $x$. Then $$f(x) =f(0)=F(\pi\mid y) =2K(y)$$