Locus and Its Equation: Derive Curves from Conditions

A locus is the set of all points satisfying a given geometric condition — for example "all points 5 cm from a fixed point" traces a circle. The equation of the locus is the algebraic relation in x and y that exactly those points satisfy. Finding it is the bridge between a verbal geometric description and an equation you can analyse.

This matters because almost every curve in coordinate geometry — circles, ellipses, parabolas, perpendicular bisectors — is defined as a locus. The standard exam method (let P = (x, y), write the condition as an equation, simplify) is identical in every case, so mastering it once unlocks a huge family of problems, including the famous Apollonius circle.

Key Concepts

• Method: let the moving point be P(x, y); express the given condition using distances, slopes or ratios; simplify to get the equation.
• Distance tool: PA = sqrt((x − xA)² + (y − yA)²); squaring early avoids radicals.
• Apollonius circle: the locus of points with PA/PB = k (k ≠ 1, constant) is a circle; for k = 1 it degenerates to the perpendicular bisector of AB.
• Verification: a claimed locus equation must contain at least one known satisfying point and exclude a known failing one.

Formulas

• Distance from P(x, y) to A(xA, yA): PA² = (x − xA)² + (y − yA)².
• Constant-ratio locus PA = k·PB gives (x − xA)² + (y − yA)² = k²[(x − xB)² + (y − yB)²], which rearranges to a circle for k ≠ 1.
• Units: the locus equation is dimensionless once coordinates carry length units consistently.
• Assumption: squaring is valid for distances since both sides are non-negative; always re-check points on the final curve.

Worked Example

Find the locus of P such that PA : PB = 1 : 2, where A = (0, 0) and B = (6, 0).

1. Let P = (x, y). The condition PA = (1/2)PB becomes 2·PA = PB.
2. Square: 4·PA² = PB², i.e. 4(x² + y²) = (x − 6)² + y².
3. Expand: 4x² + 4y² = x² − 12x + 36 + y².
4. Bring all terms left: 3x² + 3y² + 12x − 36 = 0; divide by 3: x² + y² + 4x − 12 = 0.
5. Complete the square: (x + 2)² + y² = 16. The locus is a circle with centre (−2, 0) and radius 4.
6. Verify: P = (2, 0) lies on it ((4)² + 0 = 16); PA = 2, PB = 4, ratio 1:2. Correct.

Common Misconceptions

• "The locus of equal distances is a line segment." Equidistance from two points gives the whole perpendicular bisector line, infinite in both directions.
• "Squaring always introduces fake points." For distance equations both sides are non-negative, so squaring is safe; still, verify with a test point.
• "PA/PB constant always gives a circle." Only for k ≠ 1; when k = 1 the result is a straight line, not a circle.

Real-World Applications

• GPS and trilateration: each satellite gives a constant-distance locus (a sphere); intersecting loci pin down the receiver's position.
• Antenna design: parabolic reflectors are built from the focus–directrix locus so that all reflected rays arrive in phase.
• Navigation charts: a constant-bearing locus (a rhumb line) and constant-range circles guide course plotting.
• Optics: elliptical mirrors exploit the constant-sum-of-distances locus to focus light from one focus to the other.

In short: a locus problem is solved by naming P(x, y), writing the condition algebraically, and simplifying — usually by squaring distances. The constant-ratio condition produces the Apollonius circle, and the same three-step routine generates every conic's equation.