Locus

Tie a goat to a peg in the ground with a rope that is 5 metres long. The goat wanders around, eating grass. What is the shape traced out by every point the goat can reach?

You already know the answer: a circle centred at the peg with radius 5 metres. The goat can reach any point that is exactly 5 metres from the peg, and no point that is farther. The boundary of its grazing area — the circle — is the set of all points at distance 5 from a fixed point.

That set of points has a name. It is called a locus.

The word comes from Latin, meaning "place." A locus is a place in the plane — not a single point, but an entire collection of points — defined by a rule. "All points at distance 5 from the peg" is a rule. The locus of that rule is a circle.

Here is why this idea matters. In coordinate geometry, every curve is secretly a locus. A circle is the locus of points equidistant from a centre. A parabola is the locus of points equidistant from a fixed point (the focus) and a fixed line (the directrix). A straight line is the locus of points satisfying a linear relation between x and y. Learning to find the equation of a locus is learning to build the equation of any curve from a geometric description.

From geometric rule to algebraic equation

The power of coordinate geometry is that you can take a geometric sentence — "all points equidistant from A and B" — and convert it, step by step, into an equation in x and y. Once you have the equation, you can plot it, differentiate it, find intersections with other curves, and do everything algebra allows.

Here is the general method. It works for every locus problem you will encounter.

Step 1. Let P(x, y) be an arbitrary point on the locus. You do not know what x and y are — that is the whole point. You are naming the unknown coordinates so you can set up an equation.

Step 2. Write the geometric condition in terms of x and y. If the condition says "distance from P to A equals distance from P to B," write out the distance formula for each side. If it says "the angle at P is 90°," use the slope condition for perpendicularity. Translate the English (or the diagram) into algebra.

Step 3. Simplify the equation. Square both sides if there are square roots. Expand. Cancel. Collect terms. The goal is a clean equation in x and y — ideally one you recognise as a line, a circle, a parabola, or some other familiar curve.

Step 4. Verify. Pick a point that obviously belongs to the locus (from the geometry), and check that it satisfies the equation. Pick a point that obviously does not belong, and check that it fails.

That is the entire method. The difficulty in locus problems is never the method — it is always the algebra in Step 3. The method itself is the same every time: name the point, write the condition, simplify.

The simplest locus: equidistant from two points

Start with the most basic locus problem. Given two fixed points A(1, 2) and B(5, 4), find the locus of a point P that is equidistant from A and B.

Before computing, think about what the answer should be. The set of all points equidistant from two fixed points is the perpendicular bisector of the segment joining them. That is a fact from pure geometry — every point on the perpendicular bisector is equidistant from the two endpoints, and no other point has that property. So the locus should be a straight line, perpendicular to AB and passing through the midpoint of AB.

Now derive it.

Let P(x, y) be a point on the locus. The condition is PA = PB.

Set them equal: PA = PB. Square both sides to remove the square roots (squaring is safe here because both sides are non-negative distances):

Expand each side:

The x^2 and y^2 terms appear on both sides — cancel them:

Collect terms:

That is the equation of the locus. It is a straight line, just as the geometric intuition predicted.

Verification. The midpoint of A(1,2) and B(5,4) is M(3, 3). Does M satisfy 2x + y = 9? Check: 2(3) + 3 = 9. Yes. The slope of AB is (4-2)/(5-1) = 1/2. The slope of the locus line 2x + y = 9 is -2/1 = -2. The product of slopes is (1/2)(-2) = -1, confirming perpendicularity. Everything checks out.

Notice how the x^2 and y^2 terms cancelled. That always happens in an equidistant-from-two-points problem, and it is why the result is always a line (degree one), never a curve (degree two or higher). The squared terms are the same on both sides because both distance expressions have the same x^2 + y^2 at their core.

The formal definition

With a concrete example done, here is the precise statement.

Two things to note about this definition.

First, the word "all" is doing heavy lifting. A locus is not some points that satisfy the condition — it is every point that satisfies it. The equation of a circle x^2 + y^2 = 25 is not a claim about any particular point; it is a complete description of every point on the circle.

Second, the equation must work in both directions. Every point on the locus satisfies the equation, and every point off the locus does not. If your equation is satisfied by extra points that don't belong to the locus, you have introduced an extraneous curve — usually by squaring both sides carelessly. Always verify.

Worked examples

Standard locus problems and their results

The method is always the same — name the point, write the condition, simplify — but certain conditions appear again and again. Here are the most important ones, and the curves they produce.

1. Fixed distance from a point. Condition: PA = r, where A is a fixed point and r > 0. Locus: a circle with centre A and radius r. You derived this in Example 1.

2. Equidistant from two fixed points. Condition: PA = PB. Locus: the perpendicular bisector of segment AB. A straight line. You derived this in the section above.

3. Fixed ratio of distances from two points. Condition: PA/PB = k, where k > 0 and k \neq 1. Locus: a circle (the Apollonius circle). You derived this in Example 2. When k = 1, the circle degenerates into the perpendicular bisector.

4. Equidistant from a point and a line. Condition: the distance from P to a fixed point F equals the perpendicular distance from P to a fixed line \ell. Locus: a parabola with focus F and directrix \ell. This is the definition of a parabola — you will see the full derivation in the article on Conic Sections.

5. Sum of distances from two fixed points is constant. Condition: PA + PB = 2a (where 2a > distance between A and B). Locus: an ellipse with foci at A and B. The derivation is in Ellipse.

6. Difference of distances from two fixed points is constant. Condition: |PA - PB| = 2a (where 2a < distance between A and B). Locus: a hyperbola with foci at A and B.

Every one of these can be derived using the same four-step method. The shapes are different because the conditions are different — but the process of finding the equation is identical every time.

A parametric approach: when the point depends on a parameter

Not every locus problem gives you a distance condition. Sometimes the problem says: "A point P has coordinates that depend on some parameter t. As t varies, what curve does P trace out?"

Here the approach is different. You are given x and y each as a function of t, and you need to eliminate t to find a direct relation between x and y.

Take this problem: a point P has coordinates (2t, t^2) for all real values of t. Find the equation of the locus.

Set x = 2t and y = t^2. From the first equation, t = x/2. Substitute into the second:

The locus is y = x^2/4 — a parabola, opening upward, with vertex at the origin.

Verification. At t = 3: P = (6, 9). Check: 9 = 36/4 = 9. Correct. At t = -2: P = (-4, 4). Check: 4 = 16/4 = 4. Correct.

Here is another, slightly trickier. A point P has coordinates (a\cos\theta, b\sin\theta) as \theta varies from 0 to 2\pi, with a = 5 and b = 3. Find the equation of the locus.

Set x = 5\cos\theta and y = 3\sin\theta. Then \cos\theta = x/5 and \sin\theta = y/3. Use the identity \cos^2\theta + \sin^2\theta = 1:

The locus is an ellipse with semi-major axis 5 (along the x-axis) and semi-minor axis 3 (along the y-axis). This is the standard parametric representation of an ellipse, and the elimination of the parameter \theta is exactly how you go from the parametric form to the Cartesian equation.

Applications: loci in real problems

Locus thinking appears everywhere — not just in textbook exercises, but in real-world problems and in competition mathematics.

Cell tower coverage. A cell tower at location T covers all phones within a radius r. The boundary of the coverage area is a circle — the locus of points at distance r from T. If two towers need to share coverage, the handoff boundary is the locus of points equidistant from both towers — the perpendicular bisector of the segment joining them. The entire theory of Voronoi diagrams, used in network planning across India's telecom infrastructure, rests on this locus.

Satellite dishes. A parabolic dish antenna has the property that incoming parallel signals all reflect to a single point — the focus. This works because the dish is shaped as a parabola, which is the locus of points equidistant from the focus and the directrix. The equal-distance property is what makes all the reflected paths converge. The same principle is at work in the parabolic reflectors of car headlights, where a bulb at the focus sends light out in parallel rays.

Navigation (LORAN). Before GPS, ships determined their position using LORAN — a system based on measuring the difference in arrival times of signals from two fixed radio stations. A constant time difference means a constant distance difference, and the locus of constant distance difference from two fixed points is a hyperbola. By using two pairs of stations, a ship could find two hyperbolas; its position was where they crossed.

Common confusions

• "The locus is a point." Rarely. A locus is almost always a curve (a line, a circle, a conic). The only time a locus reduces to a single point is when the condition is so restrictive that only one point satisfies it — for example, "equidistant from three non-collinear points" gives the circumcentre, a single point. But that is the exception.

• "I found the equation, so I'm done." Not quite. You should always verify that the equation does not include extra points introduced by squaring. If you squared both sides to eliminate a square root, check that every solution of the final equation actually satisfies the original condition. Squaring PA = 2 \cdot PB gives the same result as squaring PA = -2 \cdot PB (which is impossible since distances are non-negative), so in distance problems you are usually safe. But if the original condition involves directed distances or signed quantities, extraneous solutions can creep in.

• "The locus is always a conic section." It isn't. The locus can be any curve at all — a straight line, a circle, a higher-degree curve, or even a curve you don't have a name for. The standard problems in your textbook happen to produce conics because those are the curves the syllabus is building toward. But the method works for any geometric condition, regardless of what shape the answer turns out to be.

• Confusing the locus with the path of a moving point. In most textbook problems, these coincide. But technically, a locus is the set of all positions satisfying a condition — it does not require any notion of "movement." The goat tied to a peg is not tracing a circle in any particular order; the circle exists as the complete set of endpoints, all at once. This distinction matters in more advanced mathematics (topology, for instance), even if it rarely matters in coordinate geometry problems.

Going deeper

Deriving the parabola as a locus

Take the standard parabola problem: find the locus of a point P equidistant from the point F(0, a) and the line y = -a, where a > 0.

Let P(x, y) be on the locus. The distance from P to F(0, a) is:

The perpendicular distance from P(x, y) to the line y = -a is:

Set them equal:

Square both sides:

Expand both sides:

Cancel y^2 and a^2 from both sides:

This is the standard equation of a parabola with vertex at the origin, focus at (0, a), and directrix y = -a. The locus approach has just derived the equation from pure geometry — no formula was assumed, only the equal-distance condition.

For a concrete case, set a = 2. The focus is at (0, 2), the directrix is y = -2, and the equation is x^2 = 8y.

Verification at P(4, 2). Distance from P to F(0,2): \sqrt{16 + 0} = 4. Distance from P to the directrix y = -2: |2 - (-2)| = 4. Equal. The point satisfies the locus condition.

The Apollonius circle — a closer look

In Example 2 you found that the locus of PA:PB = 2:1 with A(1,0) and B(4,0) is the circle (x-5)^2 + y^2 = 4. Here is the general result.

Given two fixed points A and B separated by distance d, and a ratio k:1 with k > 0 and k \neq 1, the locus of points P with PA:PB = k:1 is a circle. Its centre lies on the line AB, and there is a clean formula for its radius.

The two points where the locus meets line AB divide segment AB internally and externally in the ratio k:1. If A and B have coordinates such that A is at the origin and B is at (d, 0), the internal division point is at \left(\frac{kd}{k+1}, 0\right) and the external division point is at \left(\frac{kd}{k-1}, 0\right).

For k = 2 and d = 3 (the distance from A(1,0) to B(4,0) in Example 2): shifting to A at origin, the internal point is at \frac{2 \cdot 3}{3} = 2 from A, i.e., at (3, 0) in the original coordinates. The external point is at \frac{2 \cdot 3}{1} = 6 from A, i.e., at (7, 0). The centre of the Apollonius circle is the midpoint of these two points: ((3+7)/2, 0) = (5, 0), and the radius is half the distance between them: (7-3)/2 = 2. This matches the equation (x-5)^2 + y^2 = 4 found in Example 2.

The Apollonius circle is one of the oldest constructions in geometry — it appears in Book VI of the Collections of Pappus, building on ideas that go back to Apollonius of Perga. The same construction appears in Indian mathematics through the study of harmonic division, and in modern physics through problems involving inverse-square-law fields.

Loci as level sets

There is a more general way to think about loci. Given any function f(x, y) of two variables, the set of points where f(x, y) = c for some constant c is called a level set (or level curve) of f. Every locus you have seen in this article is a level set:

• The circle (x-3)^2 + (y+1)^2 = 16 is the level set of f(x,y) = (x-3)^2 + (y+1)^2 at value 16.
• The line 2x + y = 9 is the level set of f(x,y) = 2x + y at value 9.
• The parabola x^2 = 8y is the level set of f(x,y) = x^2 - 8y at value 0.

Thinking of curves as level sets is central to multivariable calculus and to the study of surfaces and manifolds. The gradient of f at any point on the level set is perpendicular to the curve at that point — a fact that connects loci directly to the theory of derivatives.

Where this leads next

The locus idea is the foundation of everything that follows in coordinate geometry. Every curve you will study is a locus defined by a geometric condition.

• Straight Line — Forms — the locus of points satisfying a linear equation. Different forms for different starting information.
• Circle — Standard Forms — the locus of points at a fixed distance from a centre. The general equation, and how to read centre and radius from it.
• Conic Sections — Introduction — parabolas, ellipses, and hyperbolas as loci. The focus-directrix definition and the eccentricity parameter that unifies all three.
• Transformation of Axes — how shifting or rotating the coordinate system changes the equation of a locus without changing the curve itself.
• Coordinate Geometry Basics — the distance formula, section formula, and area formula that underpin every locus derivation.