Circles in the Coordinate Planes

Why Circles Need a Starting Point and Direction

A circle in space is determined geometrically by its center, its radius, and the plane in which it lies. A parametrization describes more: it tells us where the motion begins and which way the parameter moves around the circle. These details matter when a problem specifies a starting point or direction.

For a circle parallel to one of the coordinate planes, one coordinate remains constant. The other two coordinates vary sinusoidally. The main care needed is to keep the chosen direction consistent, especially in the \(xz\)-plane.

Coordinate Planes and Direction Convention

A circle parallel to the \(xy\)-plane has constant \(z\)-coordinate. A circle parallel to the \(yz\)-plane has constant \(x\)-coordinate, and a circle parallel to the \(xz\)-plane has constant \(y\)-coordinate.

To describe direction consistently, use these cyclic orders for the two varying coordinates:

\(xy\)-plane: \(a=x,\ b=y\), with positive direction viewed from the positive \(z\)-axis.
\(yz\)-plane: \(a=y,\ b=z\), with positive direction viewed from the positive \(x\)-axis.
\(zx\)-plane: \(a=z,\ b=x\), with positive direction viewed from the positive \(y\)-axis.

In each case, increasing angle from the first listed coordinate toward the second is the positive direction. This cyclic order is useful because it gives each coordinate plane a consistent orientation. In particular, for the \(xz\)-plane we use the order \(z,x\), not \(x,z\).

Definition: A circle of radius \(r>0\), centered at \(\mathbf{C}\) and parallel to a coordinate plane, is the set of points at distance \(r\) from \(\mathbf{C}\) whose coordinate perpendicular to that plane is fixed. A parametrization specifies a starting point and the direction in which the circle is traversed.

A Formula That Starts at Any Point

Let \(c_a,c_b\) be the center coordinates in the cyclic order for the plane, and let \(c_f\) be the fixed coordinate. Suppose the starting point has varying coordinates \(a_0,b_0\), and lies on a circle of radius \(r>0\). Define the normalized starting offsets

u=\frac{a_0-c_a}{r}, \qquad v=\frac{b_0-c_b}{r}.

Because the starting point is on the circle, \(u^2+v^2=1\). Let \(\varepsilon=1\) for the positive direction and \(\varepsilon=-1\) for the opposite direction. Then a full traversal, starting at the specified point, is given in the varying coordinates by

a(t)=c_a+r\bigl(u\cos t-\varepsilon v\sin t\bigr), \qquad b(t)=c_b+r\bigl(\varepsilon u\sin t+v\cos t\bigr), \qquad 0\le t\le 2\pi.

The fixed coordinate remains \(c_f\). The parameter \(t\) is measured in radians. The interval from \(0\) to \(2\pi\) makes one complete traversal; a shorter interval gives the corresponding arc.

Why the Formula Works

The varying coordinates are obtained by rotating the starting offset \((ru,rv)\) through the angle \(\varepsilon t\). The cosine and sine angle-addition identities give

\begin{aligned} ru\cos(\varepsilon t)-rv\sin(\varepsilon t) &=r\bigl(u\cos t-\varepsilon v\sin t\bigr),\\ ru\sin(\varepsilon t)+rv\cos(\varepsilon t) &=r\bigl(\varepsilon u\sin t+v\cos t\bigr). \end{aligned}

Here we used \(\cos(\varepsilon t)=\cos t\) and \(\sin(\varepsilon t)=\varepsilon\sin t\), since \(\varepsilon\) is either \(1\) or \(-1\). At \(t=0\), the offsets are \(ru\) and \(rv\), so the parametrization starts at \((a_0,b_0)\). At any \(t\), the squared distance from the center in the plane is

\begin{aligned} (a-c_a)^2+(b-c_b)^2 &=r^2\left[(u\cos t-\varepsilon v\sin t)^2 +(\varepsilon u\sin t+v\cos t)^2\right]\\ &=r^2(u^2+v^2)(\cos^2t+\sin^2t)\\ &=r^2. \end{aligned}

In the second line, the mixed terms cancel and \(\varepsilon^2=1\). Thus every point stays at radius \(r\). As \(t\) increases from \(0\) to \(2\pi\), the angle changes by one full turn, with sign determined by \(\varepsilon\). This proves that the formula traces exactly the required circle, once, in the chosen direction.

Worked Examples in the Coordinate Planes

Worked Example: A Circle in the \(xy\)-Plane

Find a full parametrization of the circle centered at \((2,-1,4)\), with radius \(3\), starting at \((2,2,4)\), and moving in the positive direction viewed from the positive \(z\)-axis.

The \(z\)-coordinate is fixed at \(4\). In the \(xy\)-plane, \(a=x\) and \(b=y\). The starting offsets are \(a_0-c_a=2-2=0\) and \(b_0-c_b=2-(-1)=3\), so \(u=0/3=0\), \(v=3/3=1\), and \(\varepsilon=1\). Substitution gives

\begin{aligned} x(t)&=2+3(0\cos t-1\sin t)=2-3\sin t,\\ y(t)&=-1+3(1\sin t+1\cos t) \quad \text{(not the formula's second component).} \end{aligned}

For the second component, the formula is \(b=c_b+r(\varepsilon u\sin t+v\cos t)\). Therefore the correct computation is

\begin{aligned} y(t)&=-1+3(1\cdot0\cdot\sin t+1\cdot\cos t)=-1+3\cos t,\\ z(t)&=4. \end{aligned}

At \(t=0\), these coordinates are \((2,-1+3,4)=(2,2,4)\), the required start. For every \(t\), the squared distance from the center is \((-3\sin t)^2+(3\cos t)^2=9\). Also \(x'(0)=-3\), \(y'(0)=0\): from the top of the circle, the motion initially goes left, as required for positive direction viewed from \(+z\).

Worked Example: A Circle in the \(xz\)-Plane

Parametrize the circle centered at \((1,-2,5)\), with radius \(2\), starting at \((3,-2,5)\), and moving in the positive direction viewed from the positive \(y\)-axis.

The fixed coordinate is \(y=-2\). The cyclic order is \(a=z,\ b=x\), so \(c_a=5,\ c_b=1,\ a_0=5,\ b_0=3\). Hence \(u=(5-5)/2=0\), \(v=(3-1)/2=1\), and \(\varepsilon=1\). Then

\begin{aligned} z(t)&=5+2(0\cos t-1\sin t)=5-2\sin t,\\ x(t)&=1+2(1\cdot0\cdot\sin t+1\cdot\cos t)=1+2\cos t,\\ y(t)&=-2. \end{aligned}

At \(t=0\), \((x,y,z)=(1+2,-2,5)=(3,-2,5)\). The fixed coordinate is correct, and \((x-1)^2+(z-5)^2=4\cos^2t+4\sin^2t=4\), so the radius is \(2\). The derivative at the start is \((x'(0),y'(0),z'(0))=(0,0,-2)\); in cyclic order \((z,x)\), this is motion toward decreasing \(z\) from the positive \(x\)-side, the positive direction under the stated convention.

Worked Example: A Circle in the \(yz\)-Plane

Find a full parametrization of the circle centered at \((0,1,-2)\), with radius \(4\), starting at \((0,1,-6)\), and moving in the positive direction viewed from the positive \(x\)-axis.

The fixed coordinate is \(x=0\). Here \(a=y,\ b=z\). The starting offsets are \(a_0-c_a=1-1=0\) and \(b_0-c_b=-6-(-2)=-4\). Thus \(u=0\), \(v=-1\), and \(\varepsilon=1\). The formula gives

\begin{aligned} y(t)&=1+4(0\cos t-1(-1)\sin t)=1+4\sin t,\\ z(t)&=-2+4(1\cdot0\cdot\sin t+(-1)\cos t)=-2-4\cos t,\\ x(t)&=0. \end{aligned}

At \(t=0\), the point is \((0,1,-2-4)=(0,1,-6)\). Its squared distance from the center is \((4\sin t)^2+(-4\cos t)^2=16\), so the radius is \(4\). At the start, \(y'(0)=4\) and \(z'(0)=0\): from the negative \(z\)-side, the motion initially heads toward positive \(y\), consistent with the positive cyclic direction \((y,z)\).

Choosing and Checking the Direction

The sign \(\varepsilon\) changes the direction without changing the circle or the starting point. A reliable procedure is to identify the plane's cyclic coordinate order, compute the normalized starting offsets, and then choose \(\varepsilon\) from the requested direction. Do not decide the sign from a memorized formula without checking which way the coordinates move at \(t=0\).

Identify the plane.
Record the fixed coordinate and use the cyclic order for the two varying coordinates.

Check the starting point.
Confirm that its fixed coordinate matches the plane and that its distance from the center is \(r\).

Compute \(u\) and \(v\).
Divide each starting offset from the center by the radius; check that \(u^2+v^2=1\).

Set the direction and verify.
Use \(\varepsilon=1\) or \(-1\), substitute into the formula, and check the start, radius, and initial direction.

Common Pitfall: Reversing the \(xz\)-Plane Direction

A frequent error is to treat the \(xz\)-plane as if its cyclic order were \((x,z)\), even when the direction is specified as viewed from \(+y\). For the circle from the second example, the formula \(x=1+2\cos t,\ z=5+2\sin t\) has the right center, radius, and starting point. But \(z'(0)=2\), so it moves toward increasing \(z\) at the start. Under the convention used here, that is the opposite of the positive direction viewed from \(+y\).

The correction is to use cyclic order \((z,x)\). For the required positive direction, the parametrization is \(x=1+2\cos t,\ z=5-2\sin t\). Now \(z'(0)=-2\), which gives the required initial motion. A parametrization can trace the right geometric circle and still have the wrong direction.

Why These Parametrizations Are Useful

These formulas turn geometric instructions into explicit vector-valued functions. The fixed coordinate makes the plane clear, the sine and cosine terms keep the distance from the center constant, and the initial offsets encode the starting point. The sign of the angle controls orientation. Together, these features make it possible to compare parametrizations by checking what they actually do, rather than relying on their appearance alone.

Key takeaway: To parametrize a circle parallel to a coordinate plane, keep the perpendicular coordinate fixed and rotate the starting offset in the two varying coordinates. Use the cyclic order \(xy\), \(yz\), or \(zx\) to interpret positive direction consistently.

Check Your Understanding

Use the cyclic direction convention stated above. Give exact parametrizations and check the requested features.

What coordinate stays fixed for a circle parallel to the \(yz\)-plane?
Parametrize the circle centered at \((0,2,1)\), radius \(3\), starting at \((0,5,1)\), in the positive direction viewed from \(+x\).
For a circle in the \(xz\)-plane, why is the cyclic order \((z,x)\) useful when direction is specified from \(+y\)?
For the circle centered at \((2,0,0)\) of radius \(2\), starting at \((2,0,2)\) in the \(yz\)-plane, determine the normalized offsets \(u,v\).
How can you check from a proposed parametrization that it begins at the required point and travels in the requested direction?