What Is a Locus in Geometry?
A locus is the set of all points that satisfy a specific geometric condition, and nothing else. The word is Latin for "place," and its plural is loci. Two things must always be true of a locus:
Every point that satisfies the condition is on the locus (nothing is left out).
Every point on the locus satisfies the condition (nothing extra is included).
A point moving so that it always obeys a rule "traces out" the locus. Because the rule is usually a statement about distance, a locus in a plane is normally a curve or a line, and a locus in three-dimensional space is normally a surface. The set of points is written compactly as ${P : \text{condition on } P}$, read "the set of all points $P$ such that the condition holds."
Why Does a Grazing Goat Trace a Perfect Circle?
Tie a goat to a post with a rope, and the farthest it can graze is a perfectly round patch.
The edge of that patch is not a curve someone drew. It is every point exactly one rope-length from the post, and no other point. Change the rule to "stay equally far from two posts" and the boundary snaps into a straight line instead. This idea, that a shape is really a condition on points wearing a geometric disguise, is one of the most powerful in geometry. It turns "draw this figure" into "state the rule the points obey," and once you can write the rule as an equation, the whole toolkit of coordinate geometry opens up.
What Are the Most Common Loci?
Three standard loci come up again and again, and each is defined by a single distance condition.
Locus | Condition on the point $P$ | Resulting shape |
|---|---|---|
Circle | $P$ is a fixed distance $r$ from a fixed point $O$ | A circle of radius $r$, centre $O$ |
Perpendicular bisector | $P$ is equidistant from two fixed points $A$ and $B$ | A straight line through the midpoint of $AB$, at right angles to $AB$ |
Parabola | $P$ is equidistant from a fixed point (focus) and a fixed line (directrix) | A parabola |
Angle bisector | $P$ is equidistant from two fixed lines | A pair of bisecting lines |
The circle is the locus most students meet first, and the perpendicular bisector and the parabola follow the same "equidistant" logic with different reference objects.
How Do You Find the Equation of a Locus?
The method is always the same three steps. It converts a geometric rule into algebra.
Step 1. Let the moving point be $P(x, y)$.
Step 2. Write the given condition as a distance equation using the distance formula, $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Step 3. Simplify. Squaring both sides usually clears the square roots, and the result is the equation of the locus.
The Circle as a Locus
A point $P(x, y)$ stays a fixed distance $r$ from the centre $O(0, 0)$:
$$\sqrt{(x - 0)^2 + (y - 0)^2} = r$$
Square both sides:
$$x^2 + y^2 = r^2$$
That is the equation of a circle of radius $r$ centred at the origin. Every ordered pair that fits it lies on the circle, and every point on the circle fits it.
The Perpendicular Bisector as a Locus
A point $P(x, y)$ is equidistant from $A(x_1, y_1)$ and $B(x_2, y_2)$:
$$\sqrt{(x - x_1)^2 + (y - y_1)^2} = \sqrt{(x - x_2)^2 + (y - y_2)^2}$$
Squaring and cancelling $x^2$ and $y^2$ leaves a linear equation, which is a straight line, the perpendicular bisector of segment $AB$.
The Parabola as a Locus
A point $P(x, y)$ is equidistant from the focus $(0, a)$ and the directrix $y = -a$:
$$\sqrt{x^2 + (y - a)^2} = y + a$$
Squaring gives $x^2 = 4ay$, the standard equation of a parabola.
What Are the Key Properties of a Locus?
A locus behaves in predictable ways, and these properties are worth stating on their own.
It is exact, not approximate. A locus contains every qualifying point and excludes every other one; there is no "close enough."
The condition determines the dimension. One distance condition in a plane usually gives a curve; the same idea in space usually gives a surface (a sphere, for example, is the locus of points a fixed distance from a centre).
Combined conditions intersect. A point satisfying two conditions lies on both loci at once, so its position is found where the two loci cross.
A locus can be a single point or empty. If the condition can be met by only one point, the locus is that point; if no point can meet it, the locus is empty.
The equation is the locus. Once you have the algebraic equation, the set of solutions is the locus, which is why loci sit at the heart of analytic geometry.
Where Is the Idea of a Locus Used?
A locus is a working tool across mathematics, engineering, and technology.
Navigation and GPS. A receiver equidistant in signal-time from several satellites sits where those loci (spheres) intersect; the crossing point is your location.
Design and construction. Finding a spot equidistant from two roads or two towns is a perpendicular-bisector locus problem, used in siting facilities fairly.
Physics and orbits. Planetary paths are conic-section loci, and a parabola is the locus that focuses parallel rays to a point, which is why satellite dishes and headlight reflectors are parabolic.
Robotics and safety zones. The reachable region of a robot arm, or the safe distance around a machine, is a locus defined by a maximum-distance condition.
Examples of Locus
Example 1
Find the locus of a point that moves so that it is always 5 units from the origin.
Let the point be $P(x, y)$. The condition is $OP = 5$, so by the distance formula:
$$\sqrt{x^2 + y^2} = 5$$
Square both sides:
$$x^2 + y^2 = 25$$
Final answer: the locus is a circle of radius 5 centred at the origin.
Example 2
Find the locus of a point equidistant from $A(2, 0)$ and $B(6, 0)$.
Wrong path. A student reasons that "equidistant from two points" means the midpoint, so the answer is the single point $(4, 0)$.
Why it breaks. The midpoint does satisfy the condition, but it is not the only point that does. A point directly above $(4, 0)$, say $(4, 3)$, is also equally far from $A$ and $B$. A locus must include every qualifying point, not just one.
The rescue. Set the two distances equal and simplify. With $P(x, y)$:
$$\sqrt{(x - 2)^2 + y^2} = \sqrt{(x - 6)^2 + y^2}$$
Square both sides and expand: $x^2 - 4x + 4 + y^2 = x^2 - 12x + 36 + y^2$. The $x^2$ and $y^2$ cancel, leaving $-4x + 4 = -12x + 36$, so $8x = 32$ and $x = 4$.
Final answer: the locus is the line $x = 4$, the perpendicular bisector of $AB$.
Example 3
Find the locus of a point whose distance from the point $(0, 2)$ equals its distance from the line $y = -2$.
Let $P(x, y)$. Set focus-distance equal to directrix-distance:
$$\sqrt{x^2 + (y - 2)^2} = y + 2$$
Square both sides:
$$x^2 + (y - 2)^2 = (y + 2)^2$$
Expand: $x^2 + y^2 - 4y + 4 = y^2 + 4y + 4$, so $x^2 = 8y$.
Final answer: the locus is the parabola $x^2 = 8y$.
Example 4
A point moves so that its distance from $(3, 0)$ is always twice its distance from the origin. Find its locus.
Let $P(x, y)$. The condition is $PA = 2 \cdot PO$ where $A = (3, 0)$:
$$\sqrt{(x - 3)^2 + y^2} = 2\sqrt{x^2 + y^2}$$
Square both sides: $(x - 3)^2 + y^2 = 4(x^2 + y^2)$. Expand and collect: $x^2 - 6x + 9 + y^2 = 4x^2 + 4y^2$, giving $3x^2 + 3y^2 + 6x - 9 = 0$, or $x^2 + y^2 + 2x - 3 = 0$.
Final answer: the locus is the circle $x^2 + y^2 + 2x - 3 = 0$ (centre $(-1, 0)$, radius 2).
Example 5
Find the locus of the midpoint of a ladder of fixed length 10 as it slides down a wall.
Let the ladder touch the $x$-axis at $(a, 0)$ and the $y$-axis at $(0, b)$, with $a^2 + b^2 = 10^2$. The midpoint is $P(x, y) = \left(\dfrac{a}{2}, \dfrac{b}{2}\right)$, so $a = 2x$ and $b = 2y$. Substitute into $a^2 + b^2 = 100$:
$$(2x)^2 + (2y)^2 = 100 \implies 4x^2 + 4y^2 = 100 \implies x^2 + y^2 = 25$$
Final answer: the midpoint traces the circle $x^2 + y^2 = 25$ (radius 5).
Example 6
Find the locus of points equidistant from the two lines $y = x$ and $y = -x$.
A point equidistant from two intersecting lines lies on a bisector of the angle between them. The lines $y = x$ and $y = -x$ cross at the origin, and their angle bisectors are the coordinate axes themselves.
Final answer: the locus is the pair of lines $x = 0$ and $y = 0$ (the $y$-axis and $x$-axis), an angle bisector locus.
Where Do Students Trip Up on Locus Problems?
The recurring error is treating a locus as one point instead of a whole set. Students first meeting locus problems usually find one point that fits the condition, write it down, and stop, missing that a condition on distance is satisfied by an entire curve. Naming the moving point $P(x, y)$ before doing anything else forces you to describe all qualifying positions, not just the obvious one.
Mistake 1: Giving a single point instead of the full set
Where it slips in: Any "equidistant" problem, where one convenient point (often a midpoint) clearly works.
Don't do this: Answering "$(4, 0)$" for the locus of points equidistant from $(2, 0)$ and $(6, 0)$.
The correct way: Set the two distances equal for a general point $P(x, y)$ and simplify. The answer is the line $x = 4$, which contains that midpoint and infinitely many other points.
Mistake 2: Forgetting to square both sides consistently
Where it slips in: Simplifying a distance equation with a square root on each side.
Don't do this: Squaring only one side, or squaring term by term inside a root (writing $\sqrt{(x-2)^2 + y^2}$ as $(x - 2) + y$).
The correct way: Square the entire left side and the entire right side once, then expand. The square root of a sum is not the sum of the square roots.
Mistake 3: Losing points by squaring, or missing a restriction
Where it slips in: Directrix or "distance from a line" conditions, where one side must stay non-negative.
Don't do this: Squaring $\sqrt{x^2 + (y-2)^2} = y + 2$ without noting that $y + 2 \geq 0$, then keeping stray solutions.
The correct way: Square, simplify, and then check the result against the original condition.
Conclusion
A locus is the set of all points, and only those points, that satisfy a stated condition.
A circle is the locus of points a fixed distance from a centre; a perpendicular bisector is the locus of points equidistant from two points; a parabola is the locus of points equidistant from a focus and a directrix.
To find a locus equation, let the point be $P(x, y)$, write the condition with the distance formula, and simplify.
A locus is exact: it includes every qualifying point and excludes every other one.
Combined conditions are solved where two loci intersect, the idea behind GPS positioning.
To take locus problems further with a teacher, explore Bhanzu's geometry tutor or high school math tutor sessions, or browse math classes online for structured coordinate-geometry practice.
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Practice These to Solidify Your Understanding
Work through these problems in order:
Find the locus of a point that is always 3 units from the point $(1, 2)$.
Find the locus of a point equidistant from $A(0, 0)$ and $B(0, 8)$.
Find the locus of a point whose distance from the point $(4, 0)$ equals its distance from the line $x = -4$.
Answer to Question 1: $(x - 1)^2 + (y - 2)^2 = 9$, a circle of radius 3 centred at $(1, 2)$. Answer to Question 2: $y = 4$, the perpendicular bisector of $AB$. Answer to Question 3: $y^2 = 16x$, a parabola.
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