Two spheres of radii $r_1$ and $r_2$ intersect each other orthogonally. Prove that the circle formed by the intersection of the two spheres has a radius $$\frac{r_1 r_2}{\sqrt{r_1^{2} + r_2^{2}}}.$$



See below a section of the spheres, passing through their centers $A$ and $B$. They intersect each other orthogonally if radii $AC$ and $BC$ are perpendicular.

It follows that $ABC$ is a right triangle with legs $r_1$ and $r_2$. And its altitude $CH$ is the radius of the intersection circle.

enter image description here

  • $\begingroup$ How do we show that CH is perpendicular to AB? Thanks. $\endgroup$ – R_D May 27 '18 at 14:42
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    $\begingroup$ @R_D If $D$ is the other intersection, then triangles $ABC$ and $ABD$ are congruent. It follows that $\angle CBH=\angle DBH$: triangles $BCH$ and $BDH$ are then also congruent. Hence $\angle BHC=\angle BHD$. $\endgroup$ – Intelligenti pauca May 27 '18 at 15:35
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    $\begingroup$ I referred to your sketch in my answer. Hope OK. Thanks $\endgroup$ – Narasimham Sep 15 '19 at 6:54

From Aretino's sketch considering similar right triangles

$$CH= h ;\,CA= r_1;\,CB= r_2;\, \angle ACH= \angle HBC= \theta \,$$

$$ \cos \theta = \frac{h}{r_1} ;\,\sin \theta=\frac{h}{r_2} ; $$

The result is arrived at by eliminating $\theta.$


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