Spherical surface in cartesian coordinates

Question

In the context of optics the so called biconic surface is used to describe optical elements. The biconic surface has several parameters, however, I am only interesed in the case where it reduces to the form $$ z = \frac{c(x^2 + y^2)}{1 + \sqrt{1 - c^2 (x^2 + y^2)}} $$ It is said that this equation defines a spherical surface with radius $R=1/c$. Unfortunately, I do not see how this is derived. How do I get from $$ x^2 + y^2 + z^2 = R^2 $$ to the above relation?

Answer 1

Yes, it is a spherical surface with radius $R=\frac{1}{c}$.

Write the first equation as $$z=\frac{c \left(x^2+y^2\right)}{\sqrt{1-c^2 \left(x^2+y^2\right)}+1}$$ $$\sqrt{1-c^2 \left(x^2+y^2\right)}+1=\frac{c \left(x^2+y^2\right)}{z}$$ $$\sqrt{1-c^2 \left(x^2+y^2\right)}=\frac{c \left(x^2+y^2\right)}{z}-1$$ square both sides $$1-c^2 x^2-c^2 y^2=1+\frac{c^2 x^4+2 c^2 x^2 y^2+c^2 y^4-2 c x^2 z-2 c y^2 z}{z^2}$$ simplify, multiply by $z^2$ and move everything in one side $$c^2 x^4+2 c^2 x^2 y^2+c^2 x^2 z^2+c^2 y^4+c^2 y^2 z^2-2 c x^2 z-2 c y^2 z=0$$ Simplify by $c$ $$c x^4+2 c x^2 y^2+c x^2 z^2+c y^4+c y^2 z^2-2 x^2 z-2 y^2 z=0$$ collect $x^2$ and $y^2$ $$x^2 \left(c x^2+c y^2+c z^2-2 z\right)+y^2 \left(c x^2+c y^2+c z^2-2 z\right)=0$$ and finally $$\left(x^2+y^2\right) \left(c x^2+c y^2+c z^2-2 z\right)=0$$ as $x^2+y^2=0$ has no non trivial solutions we get $$c x^2+c y^2+c z^2-2 z=0$$ divide by $c$ $$x^2+y^2+z^2-\frac{2 }{c}z=0$$ Which can be written as $$x^2+y^2+\left(z-\frac{1}{c}\right)^2=\left(\frac{1}{c}\right)^2$$ which is the equation of a sphere with canter $\left(0,0,\frac{1}{c}\right)$ and radius $R=\frac{1}{c}$.

Hope this is useful