Equation Of A Chord To A Circle

The Equation of a Chord to a Circle — A Practical Guide

Ever stared at a circle drawn on paper and wondered how you could write an equation for just one of its chords? Here's the thing — it's one of those topics in coordinate geometry that sits at the intersection of algebra and visual intuition, and once it clicks, it opens up a whole way of thinking about curves and lines together. Whether you're preparing for an exam, working through a problem set, or just genuinely curious about how circles behave, understanding the equation of a chord is a skill that pays off more than you'd expect.

This guide walks through everything from the basics to the more nuanced approaches, with enough detail that you can actually use what you learn.

What Is the Equation of a Chord to a Circle

A chord is simply a straight line segment whose endpoints both lie on the circumference of a circle. The equation of a chord refers to the algebraic expression — usually in the form of a linear equation — that describes that line.

The Standard Circle and Its General Form

Most of the work starts with a circle in standard form. The equation of a circle centered at the origin with radius r is x² + y² = r²*. If the circle is shifted so its center is at point (h, k), the equation becomes (x − h)² + (y − k)² = r².

A chord lives inside this circle. Its equation is linear — something like y = mx + c* — but the specific values of m and c depend on what information you're given about the chord itself.

What Makes a Chord Different from a Tangent

It's worth pausing here, because the distinction matters. A tangent touches the circle at exactly one point. That's why a chord cuts through it at two points. That means when you solve the system of equations formed by the circle and the chord, you should get two real solutions — two intersection points. Worth adding: if you get exactly one, you're dealing with a tangent, not a chord. If you get none, the line doesn't intersect the circle at all.

Why It Matters

You might be wondering why this is worth a full article. Here's the thing — the equation of a chord isn't just an abstract exercise. It shows up in problems about loci, in proofs involving cyclic quadrilaterals, and in more advanced topics like the equations of tangents and normals to conic sections.

Building Blocks for Harder Topics

Once you understand how to find the equation of a chord, you're better equipped to handle problems involving the midpoint of a chord, the length of a chord, and even the concept of a chord of contact (where two tangents from an external point meet the circle). These ideas cascade into each other, and the chord equation is the foundation.

Real-World Context

In engineering and physics, circles and their chords appear in problems involving circular motion, optics, and structural design. The math might be abstract, but the geometric relationships are deeply practical.

How It Works

There are several ways to find the equation of a chord, depending on what information you start with. Let's go through the main scenarios.

When You Know the Two Endpoints

If the chord connects points (x₁, y₁) and (x₂, y₂) on the circle, you can use the two-point form of a line:

(y − y₁) / (y₂ − y₁) = (x − x₁) / (x₂ − x₁)

From there, simplify to get the equation in standard or slope-intercept form. This is the most straightforward case, and it works every time — as long as the two points are actually on the circle.

When You Know the Midpoint of the Chord

This is where things get elegant. If you know the midpoint (x₁, y₁) of a chord of the circle x² + y² = r²*, there's a direct formula for the chord's equation:

x x₁ + y y₁ = x₁² + y₁²*

For a circle centered at (h, k), the formula adjusts to:

(x − h)(x₁ − h) + (y − k)(y₁ − k) = (x₁ − h)² + (y₁ − k)²

Where does this formula come from? Think about it: that perpendicularity gives you the slope relationship, and from there you can derive the equation. On top of that, it relies on the fact that the line from the center of the circle to the midpoint of a chord is perpendicular to the chord itself. It's one of those results that feels almost magical when you first see it, but it follows cleanly from basic geometry.

When You Know the Slope and a Point

If you know the slope m of the chord and a point (x₁, y₁) that lies on it (and on the circle), you use the point-slope form:

y − y₁ = m(x − x₁)*

The trick here is making sure the point you use actually lies on the circle. If you're given the slope and told the line is a chord, you may need to find where it intersects the circle first, then write the equation from there.

Want to learn more? We recommend what type of tissue is avascular and how to find the circumference when you have the diameter for further reading.

When You Know the Slope and Want the Chord of Maximum Length

The longest chord in any circle is the diameter. If you're asked for a chord with a given slope that has maximum length, you're essentially looking for the diameter parallel to that slope. The equation follows the same point-slope logic, but the line must pass through the center of the circle.

Deriving the Chord Equation from First Principles

Here's a quick walkthrough of the derivation for the midpoint formula, because seeing where it comes from makes it stick better than memorizing.

Let the circle be x² + y² = r²*, and let (x₁, y₁) be the midpoint of a chord. The endpoints of the chord are (x₁ + a, y₁ + b) and (x₁ − a, y₁ − b) for some values a and b. Both endpoints satisfy the circle's equation, so:

(x₁ + a)² + (y₁ + b)² = r² (x₁ − a)² + (y₁ − b)² = r²

Subtract the second equation from the first. The a² and b² terms cancel out, and you're left with:

4x₁a + 4y₁b = 0

Which simplifies to x₁a + y₁b = 0*. Also, this tells you that the slope of the chord is −x₁/y₁ (since the direction vector of the chord is (a, b) and b/a = −x₁/y₁*). Using point-slope form with the midpoint (x₁, y₁) and this slope, you arrive at the formula x x₁ + y y₁ = x₁² + y₁²*.

Common Mistakes / What Most People Get Wrong

Confusing the Midpoint Formula with the Tangent Formula

Confusing the Midpoint Formula with the Tangent Formula

This is the single most frequent error. The equation for a chord with midpoint $(x_1, y_1)$ on the circle $x^2 + y^2 = r^2$ is $x x_1 + y y_1 = x_1^2 + y_1^2$. The equation for the tangent* at the point $(x_1, y_1)$ on that same circle is $x x_1 + y y_1 = r^2$.

They look nearly identical. The difference lies entirely in the right-hand side: the chord uses the squared distance of the midpoint from the origin ($x_1^2 + y_1^2$), while the tangent uses the squared radius ($r^2$). Since the midpoint of a chord lies inside* the circle, $x_1^2 + y_1^2 < r^2$. Worth adding: if the midpoint happens to lie on the circle, the chord degenerates to a point, the two formulas become identical, and the line becomes the tangent. Always check: is the given point strictly inside the circle (chord) or on the circumference (tangent)?

Assuming the Midpoint Formula Works for External Points

The formula $x x_1 + y y_1 = x_1^2 + y_1^2$ is derived explicitly for the midpoint* of a chord. If you are given a point $(x_1, y_1)$ outside the circle and asked for the equation of the chord of contact (the line joining the points of tangency from that external point), the formula is $x x_1 + y y_1 = r^2$—which looks exactly like the tangent formula. If you are given an external point and asked for the chord bisected* at that point, no such real chord exists. Plugging an external point into the midpoint formula yields a line equation, but that line will not intersect the circle; it represents the polar of the point, not a real chord.

Forgetting to Translate for Non-Origin Circles

When the circle is $(x-h)^2 + (y-k)^2 = r^2$, the midpoint formula becomes $(x-h)(x_1-h) + (y-k)(y_1-k) = (x_1-h)^2 + (y_1-k)^2$. A common slip is to use the untranslated formula $x x_1 + y y_1 = x_1^2 + y_1^2$ with coordinates relative to the origin, rather than coordinates relative to the center. Always shift your coordinate system mentally (or explicitly) so the center is at $(0,0)$, apply the simple formula, then shift back.

Mixing Up Slope Signs in the Derivation

In the derivation, we found $x_1 a + y_1 b = 0$, leading to a direction vector $(a, b)$ proportional to $(-y_1, x_1)$. Worth adding: this gives a slope of $b/a = -x_1/y_1$. It is easy to accidentally flip this to $-y_1/x_1$ (the slope of the radius) or drop the negative sign. Remember: the radius to the midpoint has slope $y_1/x_1$; the chord is perpendicular, so its slope is the negative reciprocal, $-x_1/y_1$.

Conclusion

The equation of a chord is never just a formula to memorize—it is a direct consequence of the circle’s symmetry. Whether you are working from two endpoints, a midpoint, a slope, or an external point, the underlying logic remains the same: a chord is the intersection of a line and a circle, and its geometry is governed by the perpendicular relationship between the radius and the chord's midpoint.

Mastering these variations means recognizing which* geometric constraint you have been given (endpoints, midpoint, slope, length) and selecting the corresponding algebraic tool. The midpoint formula $T = S_1$ (where $T$ is the tangent form and $S_1$ is the circle equation evaluated at the point) is arguably the most powerful because it packages the perpendicularity condition into a single, memorable line. But the derivation—subtracting the two endpoint equations to cancel the quadratic terms—is the universal key. If you understand why the $a^2$ and $b^2$ terms vanish, you can reconstruct any chord equation you need, even under exam pressure, without relying on rote memory.