1
$\begingroup$

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 $$\frac{r_1 r_2}{\sqrt{r_1^{2} + r_2^{2}}}.$$

$\endgroup$
2
$\begingroup$

HINT.

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

$\endgroup$
  • $\begingroup$ How do we show that CH is perpendicular to AB? Thanks. $\endgroup$ – R_D May 27 '18 at 14:42
  • 2
    $\begingroup$ @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
  • 1
    $\begingroup$ I referred to your sketch in my answer. Hope OK. Thanks $\endgroup$ – Narasimham Sep 15 '19 at 6:54
0
$\begingroup$

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 $\theta.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.