# Locus and Circles

locus is the term used to describe the path of a single moving point that obeys certain conditions. The point A above will move around the origin at a distance 2 units away, depicted by the circle. Similarly, just like a compass, the path of a pencil is a circle with centre at the point of the compasses.

This study relates to distances between a point and a line, line and another line, or even a point to a point! Use the distance and perpendicular distance formula to solve problems relating to loci (plural form of locus).

### Examples

1. Find the equation of the locus of a point P(x, y) that moves so that is always 3 units from the origin.

You may recognise this locus as a circle, centre (0, 0) radius 3 units. Its equation is given by ${ x }^{ 2 }+{ y }^{ 2 }=9$.

Alternatively, use the distance formula.

${ d }^{ 2 }={ ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 }$

Let P(x, y) be a point of the locus.

We want PO = 3 (which is the distance 'd' in the distance formula)

${ ({ x }-0) }^{ 2 }+{ ({ y }-0) }^{ 2 }=9\\ \\ { x }^{ 2 }+{ y }^{ 2 }=9$

2. Find the equation of the locus of point P(x, y) that moves so that the distance PA to distance PB is in the ratio 2:1 where A=(-3, 1) and B=(2, -2).

Let P(x, y) be a point of the locus.

$PA:PB=2:1\\ \\ \cfrac { PA }{ PB } =\cfrac { 2 }{ 1 } \\ \\ PA=2PB\\ \\ { PA }^{ 2 }={ (2PB) }^{ 2 }\\ \\ { PA }^{ 2 } =4{ PB }^{ 2 }\\ \\ { \left[ x-(-3) \right] }^{ 2 }+{ (y-1) }^{ 2 }=4\left\{ { \left( x-2 \right) }^{ 2 }+{ \left[ y-(-2) \right] }^{ 2 } \right\} \\ \\ { \left( x+3 \right) }^{ 2 }+{ \left( y-1 \right) }^{ 2 }=4\left[ { \left( x-2 \right) }^{ 2 }+{ \left( y+2 \right) }^{ 2 } \right] \\ \\ { x }^{ 2 }+6x+9+{ y }^{ 2 }-2y+1=4\left( { x }^{ 2 }-4x+4+{ y }^{ 2 }+4y+4 \right) \\ \\ { x }^{ 2 }+6x+9+{ y }^{ 2 }-2y+1=4{ x }^{ 2 }-16x+16+{ 4y }^{ 2 }+16y+16\\ \\ 3{ x }^{ 2 }-22x+3{ y }^{ 2 }+18y+22=0$

Using Geogebra App, the trajectory of the locus is shown below:

3. Find the equation of the locus of point P(x, y) that is equidistant from a fixed point A(1, -2) and fixed line with equation y=5.

## Circle as Locus

The circle with centre (a, b) and radius r, has the equation:

${(x-a)}^{ 2 }+{(y-b)}^{ 2 }={ r }^{ 2 }$

Hence, the circle with a centre at the origin has the equation:

${ x }^{ 2 }+{ y }^{ 2 }={ r }^{ 2 }$

### Examples:

1. Find the equation of the locus of a point that is always 2 units from the point (-1, 0).

This is a circle with radius 2 and centre (-1, 0). Its equation is in the form:

${ (x-a) }^{ 2 }+{ (y-b) }^{ 2 }={ r }^{ 2 }\\ \\ { (x-(-1)) }^{ 2 }+{ (y-0) }^{ 2 }={ 2 }^{ 2 }\\ \\ { (x+1 })^{ 2 }+{ y }^{ 2 }=4\\ \\ { x }^{ 2 }+2x+1+{ y }^{ 2 }=4\\ \\ { x }^{ 2 }+2x+{ y }^{ 2 }-3=0$

2. Find the radius and the coordinates of the centre of the circle:

${ x }^{ 2 }+2x+{ y }^{ 2 }-6y-15=0\\ \\ Use\quad completing\quad the\quad square\\ \\ { x }^{ 2 }+2x+{ y }^{ 2 }-6y=15\\ \\ ({ x }^{ 2 }+2x+1)+({ y }^{ 2 }-6y+9)=15+1+9\\ \\ { \left( x+1 \right) }^{ 2 }+{ \left( y-3 \right) }^{ 2 }=25\\ \\ { \left( x-(-1) \right) }^{ 2 }+{ \left( y-3 \right) }^{ 2 }={ 5 }^{ 2 }\\$