Q.Find the equation of set of points $P$ such that $PA^2 + PB^2 = 2k^2$, where $A$ and $B$ are the points $(3, 4, 5)$ and $(-1, 3, -7)$, respectively.

The Locus of a Point: From Intuition to Precision

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

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.

1. Write the condition in words: Distance $PA = 5$.
2. Translate into algebra: $\sqrt{(x-2)^2 + (y-3)^2} = 5$.
3. 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: $PA = 2 \cdot PB$.

Step 2 — Algebra:

Step 3 — Square and simplify:

Step 4 — Complete the square:

The locus is a circle with centre $(5, 0)$ and radius 2.

Common Loci You Must Know