Two spheres of radii r1 and r2 cut orthogonally. Find radius of common circle.

Nov 2017
4
0
India
Find the radius the common circle of intersection and show that it is equal to \(\displaystyle \frac{r_1 r_2}{\sqrt{r_1^{2} + r_2^{2}}}\).

When two sphere intersect they form a circle if you join all the points of intersection, that is the common circle.
See this image, the silver portion is the common circle.
https://i.stack.imgur.com/fMCUK.jpg
 
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SlipEternal

MHF Helper
Nov 2010
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1,571
Find the radius the common circle of intersection and show that it is equal to \(\displaystyle \frac{r_1 r_2}{\sqrt{r_1^{2} + r_2^{2}}}\).
The intersection of two circles is not a common circle. Do you mean the circle of largest radius that can be circumscribed by the intersection of the two circles?
 
Nov 2017
4
0
India
See this image, the silver portion is the common circle.
https://i.stack.imgur.com/fMCUK.jpg

When two sphere intersect they form a circle if you join all the points of intersection, that is the common circle.
 
Nov 2017
4
0
India
When two sphere intersect they form a circle if you join all the points of intersection, that is the common circle.
See this image, the silver portion is the common circle.
https://i.stack.imgur.com/fMCUK.jpg
 
Nov 2017
4
0
India
I've got the centre of the common circle, which looks quite messy.
 
Jun 2013
1,096
573
Lebanon
We have a right triangle with legs \(\displaystyle r_1\) and \(\displaystyle r_2\) and we are asked to find the height
 

SlipEternal

MHF Helper
Nov 2010
3,728
1,571
We have a right triangle with legs \(\displaystyle r_1\) and \(\displaystyle r_2\) and we are asked to find the height
To expound a little on what Idea is saying, let's discuss the triangle that Idea is referring to. Draw a point at the topmost point of intersection of the two spheres (the topmost point of the common circle). Draw the two radii from the circles to that point. This gives you a right triangle (because the two spheres intersect orthogonally). The legs of the triangle are of length $r_1,r_2$ and the hypotenuse is the distance between the two centers of the two circles.

The area of the triangle is $\dfrac{1}{2}r_1r_2$ because the triangle is a right triangle. We can also represent the area of the triangle as $\dfrac{1}{2}bh$ where the base is the distance between the two centers of the spheres and the height is the radius of the common circle that you are looking for. We know that $b = \sqrt{r_1^2+r_2^2}$ because it is the hypotenuse of the triangle. Now, we set these two representations for area equal and solve for $h$ which is the radius of the common circle that we are looking for.
 
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