Ever wondered why a point moving around can magically trace a perfect circle or a straight line? That’s the charm of a locus – the hidden path that geometry loves to hide.
💡 In Simple Words: A locus is the set of all positions a point can occupy while satisfying a given condition. Think of it like the trail a dog leaves when it runs exactly 5 metres away from a tree – every spot on that trail is part of the locus.
What is a Locus? – Simple Definition
The word “locus” (pronounced “loh‑kus”) comes from Latin meaning “place”. In geometry, a locus is simply a collection of points that all meet a specific rule. For example, the rule “all points that are 3 cm from point O” draws a circle around O. The rule “all points that are the same distance from line L and point P” draws a parabola.
How to read a locus definition
When a textbook says “locus of points …”, break it down:
- Given element: the fixed point, line, or shape you’re measuring from.
- Condition: the relationship (equal distance, fixed angle, etc.).
- Resulting figure: the curve or line that satisfies the condition.
Common Locus Theorems in ICSE Geometry
ICSE expects you to know a handful of classic theorems. Here’s a quick cheat‑sheet:
- Theorem 1: The locus of points at a fixed distance r from a fixed point A is a circle with centre A and radius r.
- Theorem 2: The locus of points equidistant from two fixed points A and B is the perpendicular bisector of segment AB (a straight line).
- Theorem 3: The locus of points equidistant from a fixed point F (focus) and a fixed line d (directrix) is a parabola.
- Theorem 4: The locus of points that maintain a constant angle θ with two fixed points A and B is an arc of a circle (the so‑called “angle subtended” theorem).
- Theorem 5: The locus of points whose distances from two fixed points are in a constant ratio k is an ellipse (if k ≠ 1) or a hyperbola (if k > 1 and points are on opposite sides).
Worked Example 1: Locus of a point equidistant from a fixed point and a line
Find the locus of a point P that is always the same distance from the point F(2, 3) and the line y = 0 (the x‑axis).
- Let the distance PF = d and the perpendicular distance from P(x, y) to the line y = 0 be |y|. The condition says d = |y|.
- Write the distance formula: PF = √[(x‑2)² + (y‑3)²]. So √[(x‑2)² + (y‑3)²] = |y|.
- Square both sides (removing the absolute value): (x‑2)² + (y‑3)² = y².
- Expand and simplify: (x‑2)² + y² – 6y + 9 = y² ⇒ (x‑2)² – 6y + 9 = 0.
- Rearrange: 6y = (x‑2)² + 9 ⇒ y = [(x‑2)² + 9]/6.
This equation represents a parabola opening upwards with vertex at (2, 1.5). So the required locus is a parabola.
Worked Example 2: Locus of points at a fixed distance from a point
Show that the set of all points 4 cm from the point C(‑1, 2) is a circle.
- Use the distance formula: √[(x+1)² + (y‑2)²] = 4.
- Square both sides: (x+1)² + (y‑2)² = 16.
- This is exactly the standard equation of a circle with centre (‑1, 2) and radius 4 cm.
Hence the locus is a circle.
Quick Revision Summary
| Theorem | Condition | Resulting Locus |
|---|---|---|
| 1 | Fixed distance r from a fixed point | Circle (centre = point, radius = r) |
| 2 | Equal distance from two fixed points | Perpendicular bisector (straight line) |
| 3 | Equal distance from a fixed point (focus) and a fixed line (directrix) | Parabola |
| 4 | Constant angle subtended by two fixed points | Arc of a circle |
| 5 | Constant ratio of distances to two fixed points | Ellipse (k<1) or Hyperbola (k>1) |
📝 Likely Exam Questions
- Q1. State the locus theorem for a point moving such that it stays equidistant from two given points A and B.
Answer: The locus is the perpendicular bisector of AB. - Q2. Find the equation of the locus of a point P(x, y) that is always 5 cm from the origin.
Answer: Using the distance formula, √(x² + y²) = 5 ⇒ x² + y² = 25, a circle centred at the origin with radius 5 cm. - Q3. A point moves so that its distance from the line y = 2 is always twice its distance from the point (0, 0). Write the locus equation.
Answer: Let P(x, y). Distance to line = |y‑2|, to origin = √(x² + y²). Condition: |y‑2| = 2√(x² + y²). Squaring: (y‑2)² = 4(x² + y²) ⇒ y²‑4y+4 = 4x² + 4y² ⇒ 4x² + 3y² + 4y‑4 = 0. - Q4. Explain why the locus of points that maintain a constant angle of 90° with two fixed points is a circle.
Answer: By Thales’ theorem, any angle subtended by a diameter of a circle is a right angle. Hence all points forming a 90° angle with the two fixed points lie on the circle having the segment joining the points as its diameter. - Q5. Sketch the locus of points that are 3 cm from the line x = 1 and 4 cm from the point (0, 0). (No need for exact drawing, just describe.)
Answer: The first condition gives two parallel lines x = 1 ± 3 (i.e., x = 4 and x = ‑2). The second condition gives a circle x² + y² = 16. The locus is the intersection of the circle with the two vertical lines – two points where the circle meets each line.