Locus — The Path a Point Traces

A locus is the set of all points that satisfy a given condition — the path traced by a point as it moves under a constraint. The circle is the simplest locus: all points at a fixed distance from a centre. The parabola, ellipse, and hyperbola are all loci too.

When you study locus problems, you’re really learning to translate geometric conditions into equations. That translation is the core skill.

Key Terms and Definitions

Locus: The set of all points satisfying a given geometric condition. Plural: loci.

Equation of the locus: The algebraic relationship (equation in $x$ and $y$ ) that all points on the locus satisfy.

Geometric condition: A constraint given in the problem — like “equidistant from two fixed points” or “twice as far from A as from B.”

Perpendicular bisector: The locus of points equidistant from two fixed points. It’s a straight line.

Angle bisector: The locus of points equidistant from two lines. It’s a line (or two lines, for the internal and external bisectors).

Circle: The locus of points at a fixed distance $r$ from a fixed point (the centre).

Parabola: The locus of points equidistant from a fixed point (focus) and a fixed line (directrix).

Core Method: Finding the Equation of a Locus

The standard approach is always:

Let $P(x, y)$ be a general point on the locus
Write the geometric condition algebraically (using distance formula, midpoint formula, slope conditions, etc.)
Simplify to get the equation in $x$ and $y$
Verify with a known point if possible

The distance formula is used most often: distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ .

Standard Loci to Know

Geometric condition	Locus
Fixed distance from a point	Circle
Equidistant from two points	Perpendicular bisector (line)
Equidistant from two lines	Angle bisector
Ratio of distances from two fixed points = constant	Circle (Apollonius circle)
Sum of distances from two fixed points = constant	Ellipse
Difference of distances from two fixed points = constant	Hyperbola
Equal distances from focus and directrix	Parabola
Foot of perpendicular from a fixed point to a variable line	Circle (diameter)

Solved Examples

Easy: Perpendicular bisector as locus

Q: Find the locus of a point equidistant from $A(3, 0)$ and $B(-3, 0)$ .

Let $P(x, y)$ . Condition: $PA = PB$ .

\sqrt{(x-3)^2 + y^2} = \sqrt{(x+3)^2 + y^2}

Squaring: $(x-3)^2 = (x+3)^2$

$x^2 - 6x + 9 = x^2 + 6x + 9$

$-12x = 0 \implies x = 0$

Locus: $x = 0$ (the y-axis) — the perpendicular bisector of AB. ✓

Medium: Circle as Apollonius locus

Q: A point P moves such that its distance from $A(2, 0)$ is twice its distance from $B(-1, 0)$ . Find the locus.

Let $P(x, y)$ . Condition: $PA = 2 \cdot PB$ .

$PA^2 = 4 \cdot PB^2$

$(x-2)^2 + y^2 = 4[(x+1)^2 + y^2]$

$x^2 - 4x + 4 + y^2 = 4x^2 + 8x + 4 + 4y^2$

$0 = 3x^2 + 12x + 3y^2$

$x^2 + 4x + y^2 = 0$

$(x+2)^2 + y^2 = 4$

Locus: Circle with centre $(-2, 0)$ and radius 2. This is an Apollonius circle.

Hard (JEE Main): Locus of midpoints

Q: The ends of a line segment of length 5 move on the x and y axes respectively. Find the locus of the midpoint of the segment.

Let $A = (a, 0)$ and $B = (0, b)$ with $a^2 + b^2 = 25$ (fixed length).

Midpoint $P = (x, y) = (a/2, b/2)$ , so $a = 2x$ , $b = 2y$ .

Substituting: $(2x)^2 + (2y)^2 = 25$

x^2 + y^2 = \frac{25}{4}

Locus: Circle with centre at origin and radius $5/2$ . Clean and elegant.

Exam-Specific Tips

JEE Main: Locus problems appear in Coordinate Geometry (Straight Lines + Circles sections). The Apollonius circle (locus of points in a given ratio from two fixed points) is a favourite. Complex number locus problems ( $|z - a| = r$ ) are also common — recognize that $|z - a| = r$ is a circle with centre $a$ and radius $r$ .

CBSE Class 10 & 11: Locus problems in Class 10 Circles chapter (tangent from external point, perpendicular from centre to chord). Class 11 Straight Lines and Circles cover locus formally.

Key JEE pattern: “Find the locus of the foot of the perpendicular from a fixed point to a variable line through another fixed point.” This gives a circle (Thales’ theorem: angle in semicircle = 90°).

Common Mistakes to Avoid

Mistake 1: Not squaring to remove square roots. If your condition involves distances, you’ll have square roots. Always square both sides to clear them — then check if extraneous solutions were introduced.

Mistake 2: Forgetting to simplify to standard form. After finding $x^2 + 4x + y^2 = 0$ , complete the square to identify it as a circle. Leaving it in expanded form makes it hard to identify the locus.

Mistake 3: Not verifying with a specific point. After deriving the locus equation, plug in a point that obviously satisfies the geometric condition and check it satisfies the equation. This catches setup errors.

Mistake 4: Using the wrong condition for “twice as far.” $PA = 2PB$ means $PA^2 = 4PB^2$ after squaring. Students sometimes write $PA^2 = 2PB^2$ — off by a factor.

Mistake 5: Ignoring sign when using ratio. If a point divides AB externally in ratio $m:n$ , use the external division formula. Internal and external division are different loci.

Practice Questions

Q1: Find the locus of a point equidistant from $(0, 2)$ and $(0, -2)$ .

$PA = PB \implies y^2 + (y-2)^2 = y^2 + (y+2)^2$ … Wait, let $P = (x,y)$ : $x^2 + (y-2)^2 = x^2 + (y+2)^2 \implies -4y = 4y \implies y = 0$ . Locus is the x-axis.

Q2: Find the locus of a point $P$ such that the sum $PA + PB = 10$ , where $A = (-3, 0)$ and $B = (3, 0)$ .

This is the definition of an ellipse with foci at $(\pm 3, 0)$ , sum = 10 (so $2a = 10$ , $a = 5$ ). $b^2 = a^2 - c^2 = 25 - 9 = 16$ . Equation: $\frac{x^2}{25} + \frac{y^2}{16} = 1$ .

Q3: A point moves so that it is always 3 units from the point $(1, 2)$ . Write the equation of its locus.

$(x-1)^2 + (y-2)^2 = 9$ . Circle with centre $(1,2)$ , radius 3.

Q4: If $P(x, y)$ moves so that its distance from the x-axis equals its distance from the y-axis, what is the locus?

Distance from x-axis = $|y|$ . Distance from y-axis = $|x|$ . Condition: $|y| = |x| \implies y = \pm x$ . Locus: two lines $y = x$ and $y = -x$ .

Q5: A chord of a circle subtends a right angle at the centre. If the circle has equation $x^2 + y^2 = 25$ , find the locus of the midpoint of the chord.

Let midpoint be $M = (h, k)$ . For a chord of $x^2 + y^2 = 25$ with midpoint $(h,k)$ , the perpendicular from centre has length $\sqrt{h^2+k^2}$ . Half-chord length $= \sqrt{25 - h^2 - k^2}$ . For 90° subtended at centre, the two radii and chord form an isoceles right triangle: half-chord $=$ radius/ $\sqrt{2}$ → $\sqrt{25-h^2-k^2} = 5/\sqrt{2}$ , so $h^2 + k^2 = 25/2$ . Locus: $x^2 + y^2 = 25/2$ .

FAQs

Q: Is the locus always a curve or can it be a single point? The locus is whatever set of points satisfies the condition. It could be a curve (circle, parabola, line), a single point (if only one point satisfies), two lines (angle bisectors), or even the empty set (if no real point satisfies the condition, e.g., $|PA - PB| > |AB|$ for a hyperbola-like condition).

Q: How is locus related to conics? Every conic section is defined as a locus. Ellipse: sum of distances from two foci = constant. Hyperbola: difference of distances = constant. Parabola: distance from focus = distance from directrix. Circle: distance from centre = constant.

Q: What is an Apollonius circle? When a point P divides the segment joining two fixed points A and B in a constant ratio $k : 1$ (either internally or externally), the locus of P is a circle called the Apollonius circle. When $k = 1$ (equal distances), it degenerates to the perpendicular bisector.

Advanced Techniques

Parametric approach to locus

Sometimes the moving point depends on a parameter. The strategy: express both $x$ and $y$ in terms of the parameter, then eliminate the parameter.

A variable line passes through $(1, 0)$ with slope $m$ . It intersects $y = x^2$ at two points. Find the locus of the midpoint.

Line: $y = m(x-1)$ . Substituting into $y = x^2$ : $x^2 = m(x-1)$ , so $x^2 - mx + m = 0$ .

Roots $x_1, x_2$ : $x_1 + x_2 = m$ , $x_1 x_2 = m$ .

Midpoint: $h = \frac{x_1+x_2}{2} = \frac{m}{2}$ , so $m = 2h$ .

$k = \frac{m(x_1-1) + m(x_2-1)}{2} = \frac{m(x_1+x_2-2)}{2} = \frac{m(m-2)}{2} = \frac{2h(2h-2)}{2} = h(2h-2)$ .

So $k = 2h^2 - 2h$ . Replacing $h \to x$ , $k \to y$ : $y = 2x^2 - 2x$ .

Complex number locus

In JEE, locus problems often use complex numbers. Key results:

Condition	Locus
$	z - a
$	z - a
$\arg(z - a) = \theta$	Ray from $a$ at angle $\theta$
$	z - a

Additional Practice Questions

Q6. A rod of length $L$ has its ends on the x and y axes. Find the locus of the midpoint.

Let $A = (a, 0)$ , $B = (0, b)$ with $a^2 + b^2 = L^2$ . Midpoint $(h, k) = (a/2, b/2)$ . So $a = 2h$ , $b = 2k$ . Substituting: $4h^2 + 4k^2 = L^2$ . Locus: $x^2 + y^2 = L^2/4$ — a circle of radius $L/2$ .

Q7. Find the locus of $z$ if $\text{Re}(z^2) = 4$ .

Let $z = x + iy$ . $z^2 = x^2 - y^2 + 2ixy$ . $\text{Re}(z^2) = x^2 - y^2 = 4$ . Locus: rectangular hyperbola $x^2 - y^2 = 4$ .

Q8. The foot of the perpendicular from the origin to a variable line lies on the locus $x^2 + y^2 = 9$ . Find the equation of the variable line.

The foot of perpendicular from origin to a line lies on a circle of radius 3. The line is tangent to this circle. Equation of tangent to $x^2 + y^2 = 9$ at point $(3\cos\theta, 3\sin\theta)$ : $x\cos\theta + y\sin\theta = 3$ .