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Cone-Sphere Intersection


ConeSphereIntersectionConeSphereIntersectionCurv

Let a cone of opening parameter c and vertex at (0,0,0) intersect a sphere of radius r centered at (x_0,y_0,z_0), with the cone oriented such that its axis does not pass through the center of the sphere. Then the equations of the curve of intersection are

(x^2+y^2)/(c^2)=z^2
(1)
(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=r^2.
(2)

Combining (1) and (2) gives

 (x-x_0)^2+(y-y_0)^2+(x^2+y^2)/(c^2)-(2z_0)/csqrt(x^2+y^2)+z_0^2=r^2
(3)
 x^2(1+1/(c^2))-2x_0x+y^2(1+1/(c^2))-2y_0y+(x_0^2+y_0^2+z_0^2-r^2)-(2z_0)/csqrt(x^2+y^2)=0.
(4)

Therefore, x and y are connected by a complicated quartic equation, and x, y, and z by a quadratic equation.

If the cone-sphere intersection is on-axis so that a cone of opening parameter c and vertex at (0,0,z_0) is oriented with its axis along a radial of the sphere of radius r centered at (0,0,0), then the equations of the curve of intersection are

(z-z_0)^2=(x^2+y^2)/(c^2)
(5)
x^2+y^2+z^2=r^2.
(6)

Combining (5) and (6) gives

 c^2(z-z_0)^2+z^2=r^2
(7)
 c^2(z^2-2z_0z+z_0^2)+z^2=r^2
(8)
 z^2(c^2+1)-2c^2z_0z+(z_0^2c^2-r^2)=0.
(9)

Using the quadratic equation gives

z=(2c^2z_0+/-sqrt(4c^4z_0^2-4(c^2+1)(z_0^2c^2-r^2)))/(2(c^2+1))
(10)
=(c^2z_0+/-sqrt(c^2(r^2-z_0^2)+r^2))/(c^2+1).
(11)

So the curve of intersection is planar. Plugging (11) into (◇) shows that the curve is actually a circle, with radius given by

 a=sqrt(r^2-z^2).
(12)

See also

Cone, Sphere

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References

Kenison, E. and Bradley, H. C. Descriptive Geometry. New York: Macmillan, pp. 282-283, 1935.

Referenced on Wolfram|Alpha

Cone-Sphere Intersection

Cite this as:

Weisstein, Eric W. "Cone-Sphere Intersection." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/Cone-SphereIntersection.html

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