Sphere and plane intersection calculator Print intersection of a sphere and plane calculator ▲
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:
 
              
Solved example
Sphere and plane intersection Print sphere and plane intersection summary ▲
Sphere and plane intersection
Sphere plane draw
Plane:
Ax + By + Cz + D = 0
Sphere:
(x - xs)2 + (y - ys)2 + (z - zs)2 = R2
Note: the intersection of a plane and a sphere always forms a circle in the direction of 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
Eliminating t we get: Value of t circle center
And the intersection circle center (xc , yc , zc) is at the point:
(xc = xs + At,    yc = ys + Bt,    zc = zs + Ct)
Or after substituting the value of t:
distance of intersected circle center
The distance of intersected circle center and the sphere center is:
distance of intersected circle center
distance of intersected circle center
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
Sphere and plane intersection example Print sphere and plane intersection summary ▲
Find the radius of the circle intersected by the plane x + 4y + 5z + 6 = 0 and the sphere
(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: