Cone-Sphere Intersection

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ConeSphereIntersection

ConeSphereIntersectionCurv

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

[画像:(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

[画像: (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 = [画像:(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