Why a Circle in Space Needs Two Directions
A circle does not have to lie in a coordinate plane. It can sit in any plane in three-dimensional space, including one tilted relative to the \(xy\)-, \(yz\)-, and \(xz\)-planes.
The same basic idea used for coordinate-plane circles still applies: start at the center and move around it using cosine and sine. The difference is that the two directions of motion must now be chosen inside the circle's plane. Choosing those directions carefully turns the familiar circular parametrization into a method that works in any orientation.
The Ingredients of a Circular Parametrization
Let a circle have center \(\mathbf{c}\), radius \(r>0\), and lie in a plane. Choose two perpendicular unit vectors \(\mathbf{u}\) and \(\mathbf{v}\) that lie in that plane. The vectors \(\mathbf{u}\) and \(\mathbf{v}\) give two perpendicular directions in which to move from the center.
The term \(\mathbf{c}\) locates the center. The two remaining terms specify a displacement from the center, with length contributions \(r\cos t\) and \(r\sin t\) in the directions \(\mathbf{u}\) and \(\mathbf{v}\). The vectors must be unit length so the radius of the resulting circle is exactly \(r\).
If the available perpendicular in-plane vectors are not unit vectors, normalize them first. For a nonzero vector \(\mathbf{a}\), its unit vector is \(\mathbf{a}/\lVert\mathbf{a}\rVert\). If the two vectors are already perpendicular and have the same length \(r\), they can instead be used directly as the cosine and sine coefficient vectors.
Why the Formula Gives a Circle
The formula works for two linked reasons: every parametrized point stays in the plane, and its distance from the center is always the radius. The distance calculation depends on the vectors being perpendicular and of unit length.
Worked Example: A Circle in a Plane with Normal \((1,1,1)\)
Worked Example: A Circle in a Plane with Normal \((1,1,1)\)
Parametrize the circle of radius \(1\) centered at \(\mathbf{c}=(2,-1,4)\) in the plane \(x+y+z=5\).
The center is in the plane because \(2+(-1)+4=5\). A normal vector to the plane is \(\mathbf{n}=(1,1,1)\). To find an in-plane direction, take \((1,-1,0)\): its dot product with the normal is \(1-1+0=0\), so it is perpendicular to the normal and parallel to the plane. Its length is \(\sqrt{1^2+(-1)^2+0^2}=\sqrt{2}\). Thus \(\mathbf{u}=(1,-1,0)/\sqrt{2}\).
Choose a second direction \((1,1,-2)\). Its dot product with the normal is \(1+1-2=0\), so it too is in the plane. It is perpendicular to the first direction because \(1(1)+(-1)(1)+0(-2)=0\). Its length is \(\sqrt{1+1+4}=\sqrt{6}\), giving \(\mathbf{v}=(1,1,-2)/\sqrt{6}\).
Substitute the center, radius, and unit directions into the formula:
In component form, this is \(\mathbf{r}(t)=(2+\cos t/\sqrt{2}+\sin t/\sqrt{6},\, -1-\cos t/\sqrt{2}+\sin t/\sqrt{6},\, 4-2\sin t/\sqrt{6})\).
Check the plane condition by adding the three coordinates. The center contributes \(2-1+4=5\); the cosine terms contribute \(\cos t/\sqrt{2}-\cos t/\sqrt{2}=0\); and the sine terms contribute \(\sin t/\sqrt{6}+\sin t/\sqrt{6}-2\sin t/\sqrt{6}=0\). Hence \(x+y+z=5\) for every \(t\). The displacement from the center has squared length \(\cos^2t+\sin^2t=1\), so the radius is \(1\), as required.
Finding Directions from a Plane's Normal
In many problems, the plane is specified by an equation such as \(ax+by+cz=d\). Its normal vector is \((a,b,c)\). Any vector perpendicular to this normal is parallel to the plane. Once one in-plane direction has been selected, a second perpendicular in-plane direction can be found, for example, by using a cross product with the normal and then normalizing. The essential checks are that both directions are perpendicular to the normal and to each other, and that each has length \(1\).
Worked Example: A Circle in the Plane \(2x+y-2z=2\)
Parametrize the circle of radius \(3\) centered at \(\mathbf{c}=(1,0,0)\) in the plane \(2x+y-2z=2\).
The center satisfies the plane equation because \(2(1)+0-2(0)=2\). A normal vector is \(\mathbf{n}=(2,1,-2)\). Choose \(\mathbf{a}=(1,0,1)\). Its dot product with \(\mathbf{n}\) is \(2+0-2=0\), so it gives an in-plane direction. Its length is \(\sqrt{1+0+1}=\sqrt{2}\), and hence \(\mathbf{u}=(1,0,1)/\sqrt{2}\).
Take the cross product \(\mathbf{n}\times\mathbf{a}\): \[ (2,1,-2)\times(1,0,1)=(1,-4,-1). \] This vector is perpendicular to the normal and to \(\mathbf{a}\). Its length is \(\sqrt{1+16+1}=\sqrt{18}=3\sqrt{2}\). Thus the second unit direction is \(\mathbf{v}=(1,-4,-1)/(3\sqrt{2})\).
The parametrization is therefore \(\mathbf{r}(t)=(1,0,0)+3\cos t(1,0,1)/\sqrt{2} +3\sin t(1,-4,-1)/(3\sqrt{2})\), for \(0\leq t\leq2\pi\). Equivalently, \(\mathbf{r}(t)=(1+3\cos t/\sqrt{2}+\sin t/\sqrt{2},\, -4\sin t/\sqrt{2},\, 3\cos t/\sqrt{2}-\sin t/\sqrt{2})\).
For a check, the center contributes \(2\) to \(2x+y-2z\). The cosine contribution is \(2(3\cos t/\sqrt{2})-2(3\cos t/\sqrt{2})=0\). The sine contribution is \(2(\sin t/\sqrt{2})-4\sin t/\sqrt{2}-2(-\sin t/\sqrt{2})=0\). Thus every point satisfies the plane equation. Since \(\mathbf{u}\) and \(\mathbf{v}\) are perpendicular unit vectors, the distance from the center is \(\sqrt{9\cos^2t+9\sin^2t}=3\). The radius is correct.
When the Given Directions Are Not Unit Vectors
A problem may describe the plane using two perpendicular vectors directly. If they have unequal lengths, they cannot be used unchanged as the two direction vectors in a circle formula: unequal lengths would make the cosine and sine contributions have different maximum magnitudes. Normalize each direction, then multiply both by the circle's radius.
Worked Example: Normalize Two Given Directions
Parametrize the radius-\(2\) circle centered at \(\mathbf{c}=(2,0,2)\) in the plane \(x-y+z=4\), using the perpendicular directions \(\mathbf{a}=(1,1,0)\) and \(\mathbf{b}=(-1,1,2)\).
The center belongs to the plane since \(2-0+2=4\). The plane's normal is \((1,-1,1)\). The dot products with the two directions are \(1-1+0=0\) and \(-1-1+2=0\), confirming they lie parallel to the plane. Their mutual dot product is \(1(-1)+1(1)+0(2)=0\). Their lengths are \(\lVert\mathbf{a}\rVert=\sqrt{2}\) and \(\lVert\mathbf{b}\rVert=\sqrt{6}\). Therefore \(\mathbf{u}=(1,1,0)/\sqrt{2}\) and \(\mathbf{v}=(-1,1,2)/\sqrt{6}\).
Using radius \(2\), obtain \(\mathbf{r}(t)=(2,0,2)+2\cos t(1,1,0)/\sqrt{2} +2\sin t(-1,1,2)/\sqrt{6}\), for \(0\leq t\leq2\pi\).
At \(t=0\), the point is \((2+\sqrt{2},\sqrt{2},2)\). It lies in the plane because \(2+\sqrt{2}-\sqrt{2}+2=4\). Its displacement from the center is \((\sqrt{2},\sqrt{2},0)\), whose length is \(\sqrt{2+2}=2\). At \(t=\pi/2\), the point is \((2-2/\sqrt{6},2/\sqrt{6},2+4/\sqrt{6})\). Substitution into \(x-y+z\) gives \(2-2/\sqrt{6}-2/\sqrt{6}+2+4/\sqrt{6}=4\). Its displacement is \(2\mathbf{v}\), which has length \(2\). These sample checks agree with the general plane and radius conditions.
A Common Pitfall: Using the Plane's Normal as a Circle Direction
A normal vector points perpendicular to the plane, not along it. For the plane \(x+y+z=5\), the vector \((1,1,1)\) is normal. It cannot be used as one of the circle's direction vectors. For instance, the attempted direction \(\mathbf{w}=(1,1,1)/\sqrt{3}\) has dot product \((1,1,1)\cdot\mathbf{w}=\sqrt{3}\), not \(0\). Moving from a point in the plane in that direction leaves the plane.
The correction is to choose directions whose dot products with the normal are zero. In the first example, \((1,-1,0)\) and \((1,1,-2)\) both have zero dot product with \((1,1,1)\). After normalizing them and checking that they are perpendicular to each other, they are valid directions for the circle.
Practical Checks Before Accepting a Parametrization
A proposed formula should be checked against the geometric information, not just its appearance. For a plane \(ax+by+cz=d\) and a center \(\mathbf{c}\), first confirm that the center satisfies the plane equation. Then confirm that each direction is perpendicular to the plane's normal \((a,b,c)\), that the directions are perpendicular to each other, and that each is a unit vector. Finally, substitute the formula into the plane equation and compute its distance from the center.
Check Your Understanding
Use the plane and circle conditions to answer these questions.
- What conditions must two direction vectors satisfy to use them in the standard circle parametrization?
- Why must each direction vector be perpendicular to the plane's normal?
- A direction vector has length \(4\). What must you do before using it as a unit direction?
- For a plane with normal \((1,2,0)\), is \((2,-1,3)\) parallel to the plane? Check using a dot product.
- Explain why the displacement from the center has constant length in the circle formula.