radius of circle tangent to sphere which obscures that sphere

Question

A circular disc sits tangential to the surface of a sphere, how big does it have to be to obscure that sphere from a given point of view?

Or, in two dimensions:

Given $d$ and $r$, what is $h$?

Answer 1

From similar triangles: $$ h:d=r:\sqrt{(r+d)^2-r^2}. $$

Answer 2

If the radius of the sphere is $r$ and the distanbce between the sphere and the eye is $d$ then $$\boxed{h = r\, \sqrt{ \frac{d}{2 r+d} } }$$

Why?

The small orange line above has length $k = \frac{r^2}{r+d}$

Considering the angle $\theta$ at the eye and at the contact point we have

$$\sin(\theta) = \frac{h}{\sqrt{h^2+d^2}} = \frac{k}{r} $$

which is solved for $h = \frac{d k}{\sqrt{r^2-k^2}} = \frac{r \sqrt{d}}{\sqrt{2 r+d}}$.

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Example

A unit sphere $R=1$ and an eye a distance $d=0.8$ above the surface.

$$ h = 1 \sqrt{ \frac{0.8}{2+0.8} } = \sqrt{\frac{2}{7}} = 0.534522\;\; \checkmark $$