In short

A circle on the complex plane with centre z_0 and radius r is the set of all z satisfying |z - z_0| = r. Squaring both sides gives the general form z\bar{z} - \bar{z}_0 z - z_0\bar{z} + |z_0|^2 - r^2 = 0, which can also be written as z\bar{z} + \bar{a}z + a\bar{z} + b = 0 with a \in \mathbb{C} and b \in \mathbb{R}. The centre is -a and the radius is \sqrt{|a|^2 - b}. Intersections with lines reduce to substitution and solving a quadratic.

A mobile tower at position 2 + 3i on the complex plane has a broadcast radius of 5 km. Which points can receive its signal? Exactly the points z satisfying |z - 2 - 3i| \leq 5 -- every complex number whose distance from 2 + 3i is at most 5. The boundary of that region, |z - 2 - 3i| = 5, is a circle.

That is the entire idea: a circle on the complex plane is the set of all points at a fixed distance from a fixed centre, written as a single modulus equation. From this one-line definition, you can derive a general equation, test whether a point lies inside or outside, and find where circles intersect lines.

The standard equation

Equation of a circle in the complex plane

The circle with centre z_0 and radius r > 0 is

|z - z_0| = r.

Equivalently, squaring both sides,

|z - z_0|^2 = r^2 \quad \Longleftrightarrow \quad (z - z_0)(\bar{z} - \bar{z}_0) = r^2.

Take a concrete case. The circle with centre z_0 = 2 + 3i and radius r = 2 is |z - 2 - 3i| = 2. The point z = 4 + 3i lies on it because |4 + 3i - 2 - 3i| = |2| = 2. The point z = 2 + 5i lies on it because |2 + 5i - 2 - 3i| = |2i| = 2. The point z = 3 + 3i does not lie on it because |3 + 3i - 2 - 3i| = |1| = 1 \neq 2.

Circle with centre 2 plus 3i and radius 2A complex plane showing a circle centred at the point (2, 3) with radius 2. Red dots mark the centre and three points: (4, 3) on the right, (2, 5) at the top, and (0, 3) on the left of the circle. A dashed radius line from the centre to (4, 3) is labelled r equals 2. The point (3, 3) is shown slightly inside the circle with a note that its distance is 1, not 2. ReIm01234123z₀ = 2+3ir = 24+3i2+5i0+3i3+3i|z−z₀|=1≠2
The circle $|z - (2+3i)| = 2$. The centre is at $z_0 = 2 + 3i$ and the radius is $2$. The points $4 + 3i$, $2 + 5i$, and $3i$ all lie on the circle (each at distance $2$ from the centre). The point $3 + 3i$ is inside the circle — its distance from the centre is only $1$.

Notice: the equation |z - z_0| < r describes the interior of the circle, and |z - z_0| > r describes the exterior. The circle itself is the boundary between inside and outside.

Special case: circle centred at the origin

When z_0 = 0, the equation becomes |z| = r, or x^2 + y^2 = r^2. This is the set of all complex numbers with modulus r. You encountered this in Modulus of Complex Number — the concentric circles of constant modulus are all centred at the origin.

The unit circle |z| = 1 is especially important. It contains 1, -1, i, -i, and every \cos\theta + i\sin\theta. Complex numbers on the unit circle have modulus 1, so their conjugate equals their reciprocal: \bar{z} = 1/z.

The general form

Expand |z - z_0|^2 = r^2:

(z - z_0)(\bar{z} - \bar{z}_0) = r^2
z\bar{z} - \bar{z}_0 z - z_0\bar{z} + z_0\bar{z}_0 = r^2
z\bar{z} - \bar{z}_0 z - z_0\bar{z} + |z_0|^2 - r^2 = 0

Write a = -z_0 and b = |z_0|^2 - r^2. Then the equation becomes:

z\bar{z} + \bar{a}z + a\bar{z} + b = 0

This is the general form of a circle on the complex plane.

General equation of a circle in the complex plane

z\bar{z} + \bar{a}z + a\bar{z} + b = 0

where a \in \mathbb{C} and b \in \mathbb{R}. The centre is -a and the radius is \sqrt{|a|^2 - b}. This represents an actual circle when |a|^2 > b, a point circle when |a|^2 = b, and no real curve when |a|^2 < b.

Compare this with the general equation of a line: \bar{a}z + a\bar{z} + b = 0. The circle equation has exactly one extra term: z\bar{z}, which equals |z|^2 = x^2 + y^2. A line is the limiting case of a circle as the radius tends to infinity — the z\bar{z} term vanishes (or, more precisely, dividing through by the radius and taking the limit yields a line).

Reading centre and radius from the general form of a circleA diagram with two boxes. The left box shows the general equation z z-bar plus a-bar z plus a z-bar plus b equals 0. Arrows point from a to the label centre equals negative a, and from a and b to the label radius equals square root of |a| squared minus b. The right box shows a circle on a small complex plane with the centre and radius marked. General formzz̄ + āz + az̄ + b = 0a ∈ ℂ, b ∈ ℝCentre = −aRadius = √(|a|²−b)Exists when |a|² > bReIm−arVisual
Reading the circle from its general form. Given $z\bar{z} + \bar{a}z + a\bar{z} + b = 0$, the centre is $-a$ and the radius is $\sqrt{|a|^2 - b}$. The condition $|a|^2 > b$ ensures the radius is a positive real number — otherwise, there is no circle.

Working with the general form

Here is the practical recipe. Given an equation z\bar{z} + \bar{a}z + a\bar{z} + b = 0:

  1. Read off a from the coefficient of \bar{z} (or \bar{a} from the coefficient of z).
  2. Centre = -a.
  3. Radius = \sqrt{|a|^2 - b}.

For example, take z\bar{z} + (1 - 2i)z + (1 + 2i)\bar{z} - 4 = 0. Here \bar{a} = 1 - 2i, so a = 1 + 2i — wait, check: the coefficient of \bar{z} is 1 + 2i, so a = 1 + 2i and \bar{a} = 1 - 2i. And b = -4.

Centre = -a = -(1 + 2i) = -1 - 2i.

Radius = \sqrt{|1 + 2i|^2 - (-4)} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3.

So this is a circle with centre -1 - 2i and radius 3.

Circle through three points

Three non-collinear points determine a unique circle. Given three points z_1, z_2, z_3 on the circle, the equation z\bar{z} + \bar{a}z + a\bar{z} + b = 0 must be satisfied by each. This gives three equations (each complex equation splits into a pair of real conditions, but since a has two real parameters and b has one, the three equations are just enough to solve for a and b).

The determinant form is:

\begin{vmatrix} z\bar{z} & z & \bar{z} & 1 \\ z_1\bar{z}_1 & z_1 & \bar{z}_1 & 1 \\ z_2\bar{z}_2 & z_2 & \bar{z}_2 & 1 \\ z_3\bar{z}_3 & z_3 & \bar{z}_3 & 1 \end{vmatrix} = 0

This is analogous to the determinant condition for collinearity — but with the extra column z\bar{z} that brings in the quadratic term. When the four points are collinear (and no circle passes through them), the determinant degenerates.

Intersection of a circle and a line

Where a line meets a circle is one of the most common problems in complex-plane geometry. The strategy is direct: substitute the line equation into the circle equation.

Suppose the line is \bar{a}z + a\bar{z} + c = 0 (using c for the line's constant to avoid confusion with the circle's a). From the line equation, solve for \bar{z}:

\bar{z} = \frac{-\bar{a}z - c}{a}

Substitute this into the circle equation |z - z_0|^2 = r^2. Since |z - z_0|^2 = (z - z_0)(\bar{z} - \bar{z}_0), you get a quadratic in z. Solve it. The number of intersection points depends on the discriminant:

The condition for tangency is that the perpendicular distance from the centre to the line equals the radius.

Three cases of line-circle intersection: secant, tangent, and no intersectionThree small diagrams side by side. Left: a circle with a line crossing it at two red points (secant). Middle: a circle with a line touching it at exactly one red point (tangent). Right: a circle with a line that misses it entirely (no intersection). Below each, the condition on the distance d from centre to line is shown. Secantd < r2 pointsdTangentd = r1 pointd = rNo intersectiond > r0 pointsd
The three cases of line-circle intersection. Left: the line is a secant, cutting through the circle at two points ($d < r$). Middle: the line is tangent, touching the circle at exactly one point ($d = r$). Right: the line misses the circle entirely ($d > r$). The perpendicular distance $d$ from the centre to the line determines which case you are in.

An interactive exploration

Drag the red point along the circle below. The circle has centre z_0 = 1 + i and radius 2. The readout shows the modulus |z - z_0|, which stays at 2 for every position on the circle.

Interactive circle explorerA complex plane with a circle of radius 2 centred at 1 plus i. A red draggable point moves along the circle. Readouts show the coordinates and the distance from the point to the centre. ReIm01234123z₀=1+idrag the red point around the circle
Drag the red point along the circle $|z - (1+i)| = 2$. The readout shows $|z - z_0|^2$, which stays at $4$ (so $|z - z_0| = 2$) for every point on the circle. The circle is the set of all points at exactly this distance from the centre.

Worked examples

Example 1: Find the centre and radius of the circle $z\bar{z} - (3 + 4i)\bar{z} - (3 - 4i)z + 20 = 0$

Step 1. Identify a and b.

The equation is z\bar{z} + \bar{a}z + a\bar{z} + b = 0. The coefficient of \bar{z} in the general form is a. The given equation has -(3+4i)\bar{z}, so a = -(3+4i).

The coefficient of z is -(3-4i). In the general form this is \bar{a}. Check: \bar{a} = \overline{-(3+4i)} = -(3 - 4i). Consistent.

And b = 20.

Why: match the given equation term-by-term with the general form z\bar{z} + \bar{a}z + a\bar{z} + b = 0 to read off a and b.

Step 2. Find the centre.

\text{Centre} = -a = -\bigl(-(3 + 4i)\bigr) = 3 + 4i

Why: the centre is always -a. The double negative gives a clean result here.

Step 3. Find the radius.

r = \sqrt{|a|^2 - b} = \sqrt{|{-3 - 4i}|^2 - 20} = \sqrt{9 + 16 - 20} = \sqrt{5}

Why: |a|^2 = |{-3-4i}|^2 = 3^2 + 4^2 = 25. Since 25 > 20, the radius is real and the circle exists.

Step 4. Verify with a point on the circle. The topmost point should be at 3 + (4 + \sqrt{5})i. Check: |3 + (4+\sqrt{5})i - 3 - 4i| = |\sqrt{5}\,i| = \sqrt{5}. Confirmed.

Result: Centre = 3 + 4i, radius = \sqrt{5}.

Circle with centre 3 plus 4i and radius root 5A complex plane showing a circle centred at the point (3, 4) with radius root 5 (approximately 2.24). The centre is marked with a red dot. A dashed line from the origin to the centre shows the distance |z0| equals 5. A solid red line from the centre to a point on the circle shows the radius root 5. ReIm01234123|z₀| = 53 + 4ir = √5
The circle from Example 1: centre $3 + 4i$, radius $\sqrt{5} \approx 2.24$. The dashed line from the origin to the centre has length $|z_0| = 5$. The red segment is the radius. Matching the general form $z\bar{z} + \bar{a}z + a\bar{z} + b = 0$ gives $a = -(3+4i)$, centre $= -a = 3 + 4i$, and radius $= \sqrt{25 - 20} = \sqrt{5}$.

The Pythagorean triple (3, 4, 5) makes |a|^2 = 25, giving a clean subtraction 25 - 20 = 5 under the radical.

Example 2: Find the points where the circle $|z - i| = \sqrt{5}$ meets the line $\operatorname{Re}(z) = 1$

Step 1. Translate the conditions into Cartesian form.

The circle |z - i| = \sqrt{5} has centre z_0 = i (i.e., (0, 1)) and radius \sqrt{5}. In Cartesian form: x^2 + (y - 1)^2 = 5.

The line \operatorname{Re}(z) = 1 is simply x = 1.

Why: |z - i| = \sqrt{5} means the distance from z = x + yi to i = 0 + 1 \cdot i is \sqrt{5}. And \operatorname{Re}(z) = 1 is the vertical line at x = 1.

Step 2. Substitute x = 1 into the circle equation.

1 + (y - 1)^2 = 5
(y - 1)^2 = 4
y - 1 = \pm 2
y = 3 \quad \text{or} \quad y = -1

Why: substituting the line equation into the circle equation reduces the problem to a single-variable equation — here, a perfect square giving clean roots.

Step 3. Write the intersection points as complex numbers.

z = 1 + 3i \quad \text{and} \quad z = 1 - i

Why: x = 1 for both, and y = 3 or y = -1 from Step 2.

Step 4. Verify both points lie on the circle.

|1 + 3i - i| = |1 + 2i| = \sqrt{1 + 4} = \sqrt{5}. Check.

|1 - i - i| = |1 - 2i| = \sqrt{1 + 4} = \sqrt{5}. Check.

Both are at distance \sqrt{5} from the centre i.

Result: The circle and line intersect at z = 1 + 3i and z = 1 - i.

Circle |z minus i| equals root 5 intersecting the line x equals 1A complex plane with a circle centred at (0, 1) with radius root 5 (approximately 2.24). A vertical line at x equals 1 cuts the circle at two points: (1, 3) and (1, negative 1). Both intersection points are marked with red dots. The centre of the circle is marked at (0, 1). ReIm012−112−1z₀ = ix = 11 + 3i1 − i√5√5
The circle $|z - i| = \sqrt{5}$ intersects the vertical line $x = 1$ at two points: $1 + 3i$ (top) and $1 - i$ (bottom). Both are at distance $\sqrt{5}$ from the centre $i$, confirmed by the dashed radii. The line passes through the interior of the circle, so there are exactly two intersection points.

The intersection points are symmetric about the horizontal line y = 1 (the line through the centre parallel to the real axis). This makes sense: the line x = 1 is vertical and the centre is at x = 0, so the two intersection points are reflections of each other across the horizontal through the centre.

Common confusions

Going deeper

If you know the standard form |z - z_0| = r, the general form z\bar{z} + \bar{a}z + a\bar{z} + b = 0, how to read off centre and radius, and how to find intersections with lines, you have the full working toolkit. The material below is for readers who want connections.

Circles and Apollonius

The equation \left|\dfrac{z - z_1}{z - z_2}\right| = k, where k > 0 and k \neq 1, describes a circle called the circle of Apollonius. It is the set of all points whose distances from z_1 and z_2 are in a constant ratio k.

When k = 1, the locus degenerates to a straight line — the perpendicular bisector of the segment z_1 z_2. This is the transition case: the "circle of infinite radius." You will see this locus appear naturally in the Loci in Argand Plane article.

Squaring both sides of |z - z_1|^2 = k^2 |z - z_2|^2 and expanding gives a circle equation in the general form. The centre and radius can be extracted using the recipe above.

The power of a point

For a fixed circle |z - z_0| = r and a point p not on the circle, the quantity |p - z_0|^2 - r^2 is called the power of the point p with respect to the circle. It is positive outside the circle, zero on the circle, and negative inside. In the general form, the power of p is simply p\bar{p} + \bar{a}p + a\bar{p} + b — plug the point into the left side of the circle equation.

The power of a point has a remarkable property: if any line through p intersects the circle at points z_1 and z_2, then (p - z_1)\overline{(p - z_2)} + (p - z_2)\overline{(p - z_1)} = 2(|p - z_0|^2 - r^2). This is the algebraic foundation for the radical axis and many intersection results in geometry.

Lines as degenerate circles

On the extended complex plane (the Riemann sphere, where the point \infty is adjoined), a line is a circle that passes through \infty. This elegant unification means that every result about circles on the Riemann sphere automatically includes lines as a special case. Mobius transformations w = (az+b)/(cz+d) map circles-and-lines to circles-and-lines — this is one of their defining properties.

Where this leads next

Circles and lines are the two fundamental curves of the complex plane. The next steps combine them with geometric constraints.