Chord of Circle |z| = r Joining z₁ and z₂

Equation of the chord joining two points z₁, z₂ on the circle |z|=r. When z₁→z₂, this becomes the tangent at z₁: zz̄₁ + z̄z₁ = 2r², equivalently z/z₁ + z̄/z̄₁ = 2. The chord of contact from an external point z₀ is: zz̄₀ + z̄z₀ = 2r². Connects back to: Circles chapter (chord, tangent, chord of contact).

For the circle $|z| = r$, every point satisfies $z\bar{z} = r^2$. This single identity is the key to deriving both the chord and tangent equations cleanly.

Setup

Let $z_1$ and $z_2$ be two points on $|z| = r$. So:

$$z_1\bar{z}_1 = r^2 \quad \text{and} \quad z_2\bar{z}_2 = r^2$$

Equation of the Chord

The equation of the line through $z_1$ and $z_2$ (from the general line formula) can be written as:

$$z(\bar{z}_1 + \bar{z}_2) + \bar{z}(z_1 + z_2) = 2r^2$$

Verification: Substitute $z = z_1$:

$$z_1(\bar{z}_1 + \bar{z}_2) + \bar{z}_1(z_1 + z_2) = z_1\bar{z}_1 + z_1\bar{z}_2 + \bar{z}_1 z_1 + \bar{z}_1 z_2 = 2r^2 + z_1\bar{z}_2 + \bar{z}_1 z_2$$

Wait — this equals $2r^2 + 2\operatorname{Re}(z_1\bar{z}_2)$, not $2r^2$ in general. Let me derive more carefully.

Correct derivation: For the line through two points $z_1$, $z_2$ on $|z|=r$, use the fact that any point $z$ on the chord satisfies collinearity with $z_1$, $z_2$ — i.e., $(z - z_1)/(\bar{z} - \bar{z}_1)$ equals the complex slope $(z_2 - z_1)/(\bar{z}_2 - \bar{z}_1)$.

Since $z_1\bar{z}_1 = r^2$, we have $\bar{z}_1 = r^2/z_1$. Substituting both $\bar{z}_1$ and $\bar{z}_2$:

$$\bar{z}_1 = \frac{r^2}{z_1}, \qquad \bar{z}_2 = \frac{r^2}{z_2}$$

The chord equation reduces to:

$$z \cdot \frac{r^2}{z_1 z_2}(z_1 + z_2) + \bar{z}(z_1 + z_2) = \frac{r^2(z_1+z_2)^2}{z_1 z_2} - \ldots$$

The cleanest route: use the parametric approach with $z_1 = re^{i\alpha}$, $z_2 = re^{i\beta}$.

The chord joining $re^{i\alpha}$ and $re^{i\beta}$ has equation:

$$z e^{-i(\alpha+\beta)/2} + \bar{z} e^{i(\alpha+\beta)/2} = 2r\cos\!\frac{\alpha-\beta}{2}$$

Tangent at a Point

As $z_2 \to z_1$ (i.e., $\beta \to \alpha$), the chord becomes the tangent. Setting $\alpha = \beta$:

$$z e^{-i\alpha} + \bar{z} e^{i\alpha} = 2r$$

Multiplying through by $e^{i\alpha}$ and using $z_1 = re^{i\alpha}$, so $e^{i\alpha} = z_1/r$:

$$z \cdot \frac{\bar{z}_1}{r} + \bar{z} \cdot \frac{z_1}{r} = 2r$$ $$\boxed{z\bar{z}_1 + \bar{z}z_1 = 2r^2}$$

Chord of Contact

For an external point $z_0$, the chord joining the two tangent points has equation:

$$z\bar{z}_0 + \bar{z}z_0 = 2r^2$$

This has the same form as the tangent equation — replacing the point on the circle $z_1$ with the external point $z_0$. This is the $T = S_1$ pattern in complex form, connecting back to the Circles chapter.