# Equation of Locus

A **locus** is a set of points which satisfy certain geometric conditions. Many geometric shapes are most naturally and easily described as loci. For example, a circle is the set of points in a plane which are a fixed distance \( r\) from a given point \( P,\) the center of the circle.

Problems involving describing a certain locus can often be solved by explicitly finding equations for the coordinates of the points in the locus. Here is a step-by-step procedure for finding plane loci:

Step 1:If possible, choose a coordinate system that will make computations and equations as simple as possible.

Step 2:Write the given conditions in a mathematical form involving the coordinates \(x\) and \(y\).

Step 3:Simplify the resulting equations.

Step 4:Identify the shape cut out by the equations.

Step 1 is often the most important part of the process since an appropriate choice of coordinates can simplify the work in steps 2-4 immensely.

## Examples

Find the locus of points \(P\) such that the sum of the squares of the distances from \( P\) to \( A\) and from \( P \) to \( B,\) where \(A\) and \(B\) are two fixed points in the plane, is a fixed positive constant.

After rotating and translating the plane, we may assume that \( A = (-a,0)\) and \(B = (a,0).\) Suppose the constant is \( c^2,\) \( c\ne 0.\) Then

\[\begin{align} PA^2 + PB^2 &= c^2 \\ (x+a)^2+y^2+(x-a)^2+y^2 &= c^2 \\ 2x^2+2y^2+2a^2 &= c^2 \\ x^2+y^2 &= \frac{c^2}{2}-a^2. \end{align}\]

So the locus is either empty \(\big(\)if \(c^2 < 2a^2\big),\) a point \(\big(\)if \( c^2=2a^2\big),\) or a circle \(\big(\)if \(c^2>2a^2\big).\)

Describe the locus of the points in a plane which are equidistant from a line and a fixed point not on the line.

After rotation and translation (and possibly reflection), we may assume that the point is \( (0,2a)\) with \( a\ne 0\) and that the line is the \( x\)-axis. The distance from \((x,y)\) to the \(x\)-axis is \( |y|,\) and the distance to the point is \( \sqrt{x^2 + (y-2a)^2},\) so the equation becomes

\[\begin{align} y^2 &= x^2+(y-2a)^2 \\ 0 &= x^2-4ay+4a^2 \\ y &= \frac{x^2}{4a} + a, \end{align}\]

which describes a parabola.

Note that if the point did lie on the line, e.g. \(a=0,\) the equation reduces to \( x^2=0,\) or \(x=0,\) which gives a line perpendicular to the original line through the point; this makes sense geometrically as well. \(_\square\)

Find the locus of all points \( P\) in a plane such that the sum of the distances \(PA\) and \(PB\) is a fixed constant, where \(A\) and \(B\) are two fixed points in the plane.

After translating and rotating, we may assume \( A = (-a,0)\) and \(B = (a,0),\) and let the constant be \( c.\) If \( c < 2a,\) then the locus is clearly empty, and if \( c=2a,\) then the locus is a point, so assume \( c>2a.\) Let \(PA = d_1\) and \( PB=d_2.\) Then \( d_1^2+d_2^2 = (x+a)^2+y^2+(x-a)^2+y^2 = 2x^2+2y^2+2a^2,\) and \( d_1^2-d_2^2 = 4ax.\) The locus equation is

\[\begin{align} d_1+d_2 &= c \\ d_1^2+d_2^2+2d_1d_2 &= c^2 \\ 4d_1^2d_2^2 &= \big(c^2-d_1^2-d_2^2\big)^2 \\ 4d_1^2d_2^2 &= c^4 - 2c^2\big(d_1^2+d_2^2\big) + \big(d_1^2+d_2^2\big)^2 \\ 0 &= c^4-2c^2\big(d_1^2+d_2^2\big) + \big(d_1^2-d_2^2\big)^2 \\ 0 &= c^4-2c^2\big(2x^2+2y^2+2a^2\big)+16a^2x^2 \\ \big(4c^2-16a^2\big)x^2+\big(4c^2\big)y^2 &= c^2\big(c^2-4a^2\big). \end{align}\]

Since \( 4c^2-16a^2>0\) and \( c^2-4a^2>0,\) this is the equation of an ellipse. \(_\square\)

Note that if \(a=0,\) this describes a circle, as expected \((A\) and \(B\) coincide\().\)

\(A\) and \(B\) are two points in \(\mathbb{R}^2\). What is the locus of points such that the ratio of the distances from \(A\) and \(B\) is always \(\lambda:1\), where \(\lambda\) is a positive real number not equal to \(1?\)

## See Also

**Cite as:**Equation of Locus.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/equation-of-locus/