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Sphere and plane intersection calculator

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Plane equation:
x + y + z + = 0
Plane passing through 3 points:
 xa       ya       za 
 xb       yb       zb 
 xc       yc       zc 
Sphere form:     (x - a)2 + (y - b)2 + (z - c)2 = r2
( x −  )2 + ( y −  )2 + ( z −  )2 = 
 2
Sphere form:     x2 + y2 + z2 + Ax + By + Cz + D = 0
x2 + y2 + z2 +   x +   y +   z +   = 0
Intersected circle center:
Intersected circle radius:
Intersected circle area:
Distance of sphere center to plane:
Sphere center to plane vector:
Sphere center to plane line equation:
              
Sphere plane intersection summary
Solved example

Sphere and plane intersection summary

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Sphere plane draw
General equation of a plan is:
Ax+By+Cz+D=0
General equation of a sphere is  (xs , ys , zs is the center of the sphere):
(x-x_s )^2+(y-y_s )^2+(z-z_s )^2=r^2

the intersection of a plane and a sphere always forms a circle which is perpendicular to the normal vector to the plane, and an ellipse on the projections on the x, y, z axes.

The vector normal to the plane is:   n = Ai + Bj + Ck   this vector is in the direction of the line connecting sphere center and the center of the circle formed by the intersection of the sphere with the plane. This vector when passing through the center of the sphere (xs , ys , zs)  forms the parametric line equation

L={(x,y,z): xs + At , ys + Bt , zs + Ct}

In order to find the intersection circle center, we substitute the parametric line equation into the plane equation.

A(xs + At) + B(ys + Bt) + C(zs + Ct) + D = 0
Solving  t  we find:
Value of t circle center

And the intersected circle center (xc , yc , zc) is at the point:     (xc = xs + At,    yc = ys + Bt,    zc = zs + Ct)

Substituting the value of t we can find the intersected circle center:

distance of intersected circle center

The distance of intersected circle center and the sphere center is:

d=√((x_s-x_c )^2+(y_s-y_c )^2+(z_s-z_c )^2 )

d=|Ax_s+By_s+Cz_s+D|/√(A^2+B^2+C^2 )

Intersected circle radius is:
Intersected circle radius
Condition for sphere and plane intersection:
R > d
Condition for sphere and plane tangency:
R = d
Condition for no intersection:
R < d

Example 1 − Sphere and plane intersection

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Find the radius of the circle that is intersected by the plane x + 4y + 5z + 6 = 0, the sphere is given by the equation
(x − 1)2 + (y + 1)2 + (z − 3)2 = 9
The vector normal to the plane is:
n = i + 4j + 5k
The center of the sphere is at point:
(1 , − 1 , 3)

So, the equation of the parametric line which passes through the sphere center and is normal to the plane is:

L = {(x, y, z):   x = 1 + t       y = − 1 + 4t       z = 3 + 5t}
This line passes through the circle center formed by the plane and sphere intersection, in order to find the center point of the circle we substitute the line equation into the plane equation

1 + t + 4(− 1 + 4t) + 5(3 + 5t) + 6 = 0

After solving for t we get the value:     t = − 0.43

And the circle center point is at:     (1 − 0.43 ,    − 1 − 4*0.43 ,    3 − 5*0.43) = (0.57 , − 2.71 , 0.86)

The distance of this point to the sphere center is:
And the intersected circle radius is: