Imagine you're walking in a park, but you must always stay exactly 5 metres away from a fountain at the centre. As you walk, your path traces out a circle. That circle is the locus of your position — the set of all points that satisfy the rule "distance from fountain = 5 m".
Now think of a different rule: you must always be equally far from two trees. Your path becomes the perpendicular bisector of the line joining those trees — a straight line.
Every geometric shape you know — circle, line, parabola, ellipse — is really just a locus. A circle is the set of points at a fixed distance from a centre. A line is the set of points that satisfy a linear equation. The word "locus" (plural: loci) simply means "place" or "path" in Latin.
The Precise Definition
Important
Locus of a point is the set of all positions (points) that satisfy a given geometric condition or a set of conditions.
In coordinate geometry, a locus is represented by an equation in x and y (or x, y, z in 3D). Every point (x,y) that satisfies the condition lies on the locus; every point that does not satisfy it lies off the locus.
How to Find the Equation of a Locus
The process is mechanical. Suppose a point P(x,y) moves so that its distance from a fixed point A(2,3) is always 5 units.
Write the condition in words: Distance PA=5.
Translate into algebra: (x−2)2+(y−3)2=5.
Simplify: Square both sides: (x−2)2+(y−3)2=25.
That's it. The locus is a circle with centre (2,3) and radius 5.
A Slightly Harder Example
Find the locus of a point P(x,y) that moves so that its distance from A(1,0) is twice its distance from B(4,0).
Step 1 — Condition: PA=2⋅PB.
Step 2 — Algebra:
(x−1)2+y2=2(x−4)2+y2
Step 3 — Square and simplify:
(x−1)2+y2=4[(x−4)2+y2]
x2−2x+1+y2=4(x2−8x+16+y2)
x2−2x+1+y2=4x2−32x+64+4y2
0=3x2−30x+3y2+63
x2−10x+y2+21=0
Step 4 — Complete the square:
(x2−10x+25)+y2=4
(x−5)2+y2=4
The locus is a circle with centre (5,0) and radius 2.
Tip
When the condition involves distances in a ratio, the locus is often a circle (called the Apollonius circle). If the ratio is 1:1, the locus is the perpendicular bisector — a straight line.
Common Loci You Must Know
Condition
Locus
Equation (standard form)
Fixed distance from a point
Circle
(x−h)2+(y−k)2=r2
Equal distances from two points
Perpendicular bisector
Linear equation
Fixed distance from a line
Pair of parallel lines
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Sum of distances from two fixed points is constant
The set of points is the sphere 2x2+2y2+2z2−4x−14y+4z=k2−109, i.e. (x−1)2+(y−27)2+(z+1)2=42k2−161, centred at the midpoint (1,27,−1) of AB.
Understanding the locus
We want all points P for which the sum of the squared distances to two fixed points A and B is a fixed constant k2. Expanding the two squared distances, the x2,y2,z2 terms add up (they do not cancel), so the result is a sphere — centred, it turns out, at the midpoint of AB.
Step-by-step solution
Let P=(x,y,z), with A=(3,4,5) and B=(−1,3,−7).
Step 1 — Squared distances.
PA2=(x−3)2+(y−4)2+(z−5)2=x2+y2+z2−6x−8y−10z+50,
PB2=(x+1)2+(y−3)2+(z+7)2=x2+y2+z2+2x−6y+14z+59.
Step 2 — Apply PA2+PB2=k2.
2x2+2y2+2z2−4x−14y+4z+109=k2,
so the required equation is
2x2+2y2+2z2−4x−14y+4z=k2−109.
Step 3 — Centre–radius form. Divide by 2:
x2+y2+z2−2x−7y+2z+2109=2k2.
Complete the square: x2−2x=(x−1)2−1, y2−7y=(y−27)2−449, z2+2z=(z+1)2−1. Carrying the constants across, …