Horizontal Slices of a Sphere
A sphere is a surface in three-dimensional space, but a horizontal plane can cut it in a circle. The circle’s radius depends on how far the plane is above or below the sphere’s center: slices near the center are larger, while slices near the top or bottom shrink to a point.
This gives a direct way to build a parametrization. First calculate the circle’s radius using the sphere equation; then use the standard cosine-and-sine parametrization in the horizontal plane. The same calculation also tells us when the plane misses the sphere or just touches it.
The Sphere and a Horizontal Plane
A sphere with center \((a,b,c)\) and radius \(R>0\) is the set of points \((x,y,z)\) satisfying
A horizontal plane has equation \(z=h\), where \(h\) is its height. At every point in the plane, the vertical displacement from the sphere’s center is the same number, \(h-c\). Substituting \(z=h\) into the sphere equation leaves a relation between \(x\) and \(y\):
The right-hand side is the square of the radius of the horizontal circle, when that quantity is positive.
The word “height” here means the coordinate \(z\), while the relevant vertical distance is \(h-c\). These coincide only when the sphere’s center has \(z\)-coordinate zero.
Radius and Parametrization
Define \(\rho\) to be the nonnegative square root of \(R^2-(h-c)^2\), when that quantity is nonnegative. For a positive radius, the horizontal circle has parametrization
The first two coordinates describe a circle of radius \(\rho\) centered at \((a,b)\); the last coordinate holds the curve in the plane \(z=h\). As \(t\) runs from \(0\) to \(2\pi\), the circle is traced once, with the starting point repeated at the endpoint. Using \(0\le t<2\pi\) instead avoids that repeated endpoint.
The strict inequality for a circle matters. When \(|h-c|=R\), the calculated radius is zero, so the parametrization is constant and represents a point rather than a circle.
Why the Formula Gives Exactly the Cross-Section
There are two things to check: every parametrized point lies on both the sphere and the plane, and every point in the cross-section is reached by the parametrization. The first check establishes that the parametrization stays on the desired curve; the second establishes that it does not leave part of the circle out.
The third coordinate is \(z=h\) for every parameter value, so every parametrized point lies in the horizontal plane.
The first two coordinate differences from the center are \(\rho\cos t\) and \(\rho\sin t\), while the vertical difference is \(h-c\).
Every point of the cross-section has horizontal displacements whose squares sum to \(\rho^2\). Such a point on a circle can be represented by \((\rho\cos t,\rho\sin t)\) for some angle \(t\).
More explicitly, for the parametrization we have
Here the identity \(\cos^2 t+\sin^2 t=1\) and the definition \(\rho^2=R^2-(h-c)^2\) give the last two equalities. Thus each parametrized point lies on the sphere.
For the converse, take any point \((x,y,h)\) in the cross-section. Substituting it into the sphere equation gives \((x-a)^2+(y-b)^2=\rho^2\). When \(\rho>0\), the two normalized displacements \((x-a)/\rho\) and \((y-b)/\rho\) have squares summing to \(1\). Therefore they are the cosine and sine of some angle \(t\), so the point is \((a+\rho\cos t,b+\rho\sin t,h)\). This proves that the parametrization covers the entire cross-section.
Worked Example: A Latitude Circle on a Sphere Centered at the Origin
Worked Example: A Latitude Circle on a Sphere Centered at the Origin
Find and verify the horizontal cross-section at height \(z=3\) of the sphere \(x^2+y^2+z^2=25\).
The center is \((0,0,0)\), the radius is \(R=5\), and the plane height is \(h=3\). Thus the vertical displacement from the center is \(h-c=3-0=3\). The squared horizontal radius is
Since \(\rho\) is nonnegative, \(\rho=\sqrt{16}=4\). A parametrization is
To check it, substitute the coordinates into the sphere equation:
The third coordinate is always \(3\), so the curve also lies in the required plane. The computed radius and the sphere equation agree for every \(t\).
Worked Example: A Sphere with a Shifted Center
Worked Example: A Sphere with a Shifted Center
Find and verify the intersection of the sphere centered at \((1,-2,3)\) with radius \(5\) and the plane \(z=6\).
The vertical displacement is \(h-c=6-3=3\), not \(6\). Therefore
The circle’s center in the plane is \((1,-2,6)\), so a parametrization is
For a direct check, the coordinate differences from the sphere center are \(4\cos t\), \(4\sin t\), and \(6-3=3\). Hence
This is the sphere’s equation, and the third coordinate remains \(6\), as required. The circle is centered at \((1,-2,6)\) within the plane and has radius \(4\).
Worked Example: Tangency and No Intersection
Worked Example: Tangency and No Intersection
Consider the sphere \(x^2+y^2+z^2=25\). Determine what happens when the horizontal plane has height \(z=5\), and when it has height \(z=6\).
For \(z=5\), the vertical displacement from the center is \(5\), so the squared horizontal radius is
The only point is \((0,0,5)\). Substitution verifies it: \(0^2+0^2+5^2=25\). Since the radius is zero, this is a single tangent point, not a circle.
For \(z=6\), the proposed squared radius is
A real radius cannot have a negative square. Equivalently, every point with \(z=6\) has \(x^2+y^2+z^2\ge 0+36=36>25\), so no such point lies on the sphere. The plane does not intersect the sphere.
Other Curves on a Sphere
A latitude circle is only one possible curve on a sphere. For example, the sphere \(x^2+y^2+z^2=25\) also contains the vertical circle in the plane \(y=0\), parametrized by
Indeed, \(y=0\) throughout, and
This is a circle lying on the sphere, but it is not a horizontal circle: its \(z\)-coordinate varies with \(t\). The example illustrates an important distinction. The sphere equation specifies a surface, not one particular curve. To select a curve, we need an additional description, such as a fixed height for a latitude circle or a fixed vertical plane for this example.
A Common Radius Mistake
For a sphere whose center is not at the origin, a common error is to use the plane’s height \(h\) as the vertical displacement. For the shifted sphere in the second example, this incorrect calculation would be
That answer incorrectly suggests there is no intersection. The calculation used the plane height \(6\), but the sphere center is at height \(3\). The correct displacement is \(6-3=3\), giving
The plane therefore cuts the sphere in a circle of radius \(4\). As a check, the point \((5,-2,6)\), obtained by taking \(t=0\) in the parametrization, has squared distance \((5-1)^2+(-2+2)^2+(6-3)^2=16+0+9=25\) from the center.
Why These Parametrizations Are Useful
The height-and-radius calculation turns a three-dimensional geometry question into a simple relation among the sphere’s radius, the vertical displacement, and the horizontal radius. It gives both a practical construction and a reliable test: if the proposed height is too far from the center, there is no real horizontal cross-section; if it is exactly one radius away, the cross-section is a point.
When building a parametrization, keep the center and the displacement clear. The circle’s center is \((a,b,h)\), not generally the sphere center \((a,b,c)\); its radius is based on \(h-c\), not just on \(h\). Substituting the final coordinates into the sphere equation is a direct way to verify the result.
Check Your Understanding
Use the sphere equation and the height relative to its center to answer these questions.
- What is the radius of the horizontal cross-section of \(x^2+y^2+z^2=100\) at height \(z=6\)? Write a parametrization and check it in the sphere equation.
- A sphere has center \((2,1,-1)\) and radius \(5\). At what horizontal heights does its cross-section reduce to a single point?
- For that same sphere, find the radius and give a parametrization of the cross-section at \(z=2\).
- Why does a horizontal plane at height \(z=5\) miss the sphere \(x^2+y^2+z^2=9\)?
- Give one reason the curve \((5\cos t,0,5\sin t)\) is not a latitude circle on the sphere \(x^2+y^2+z^2=25\).