General Locus Problems
Defining a locus and finding an equation
A locus is a pathway (perhaps a line or a curve) formed by a collection of points, each of which has a location that satisfies some given condition.
For example, a straight line can be defined as the pathway that consists of all points satisfying an equation of the form $y=mx+b$y=mx+b. Points that are not on this line will not satisfy this equation and so are not part of the locus.
In the diagram the red points satisfy the line $y=\frac{1}{2}x+3$y=12x+3 as exemplified by the point $\left(12,9\right)$(12,9), whereas the blue points, such as the point $\left(2,2\right)$(2,2), don't.
In the study of cartesian geometry, we are often tasked with constructing an equation of a line or a curve that is based on some imposed condition.
For example, suppose we wish to find the locus of all points $\left(x,y\right)$(x,y) that are at a distance of exactly $5$5 units from the origin. Just like a child on a play horse on the outer rim of a merry-go-round the locus becomes a circle with centre $\left(0,0\right)$(0,0) and radius $5$5.
From the distance formula, we can easily locate a point, say $\left(3,4\right)$(3,4), that satisfies the imposed condition:
| $d$d | $=$= | $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$√(x2−x1)2+(y2−y1)2 |
| $=$= | $\sqrt{\left(3-0\right)^2+\left(4-0\right)^2}$√(3−0)2+(4−0)2 | |
| $=$= | $\sqrt{9+16}$√9+16 | |
| $=$= | $\sqrt{25}$√25 | |
| $\therefore$∴ $d$d | $=$= | $5$5 |
But what about any point $\left(x,y\right)$(x,y) on the locus?
Again, using the distance formula, and setting $d=5$d=5, we have:
| $5$5 | $=$= | $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$√(x2−x1)2+(y2−y1)2 |
| $5$5 | $=$= | $\sqrt{\left(x-0\right)^2+\left(y-0\right)^2}$√(x−0)2+(y−0)2 |
| $5$5 | $=$= | $\sqrt{x^2+y^2}$√x2+y2 |
| $\therefore$∴ $x^2+y^2$x2+y2 | $=$= | $25$25 |
This is such an interesting result. For example, the left hand side of this equation is symmetrical - we could swap values for $x$x and $y$y around and it would not make any difference to the sum. Hence we know $\left(3,4\right)$(3,4) and $\left(4,3\right)$(4,3) will both be on the circle.
Because the left hand side is the sum of two squares, we also know that $\left(-3,4\right)$(−3,4), $\left(-3,-4\right)$(−3,−4), $\left(-4,3\right)$(−4,3) and $\left(-4,-3\right)$(−4,−3) satisfy the equation. The shape of the locus reflects this symmetric relationship between $x$x and $y$y.
Of course there are an infinite set of points $\left(x,y\right)$(x,y) that satisfy $x^2+y^2=25$x2+y2=25. Points like $\left(0,5\right)$(0,5), $\left(0,-5\right)$(0,−5) and $\left(\sqrt{10},\sqrt{15}\right)$(√10,√15) for example.
In general terms, the locus of all points lying a distance $r$r from the origin is given by $x^2+y^2=r^2$x2+y2=r2. Even more generally, the locus of all points lying a distance $r$r units from the fixed point $\left(h,k\right)$(h,k) is given, through the same process, by $\left(x-h\right)^2+\left(y-k\right)^2=r^2$(x−h)2+(y−k)2=r2.
Other examples
Example 1
Find the equation of the locus of all points $7$7 units from the $x$x axis.
Here the locus is simply the two lines that are parallel to the $x$x axis and a distance of $7$7 away from it. The locus includes all points on both of the lines given by $y=\pm7$y=±7.
Example 2
Sometimes we talk about a variable point $\left(x,y\right)$(x,y) that moves according to the imposed condition, just like a train moves according to where the train tracks are pointing. Here is an example:
Find the locus of a variable point which moves so that its distance from the $y$y axis is always $3$3 times its distance from the $x$x axis.
The diagram makes it clear that the value of the $x$x component must remain $3$3 times the value of the $y$y component. That is to say, for the point $\left(x,y\right)$(x,y), we have $x=3y$x=3y, and so the locus becomes the line given by $y=\frac{1}{3}x$y=13x.
Example 3
Find the locus of the point which moves such that it remains equidistant from the two fixed points $\left(5,3\right)$(5,3) and $\left(2,8\right)$(2,8).
Here, we have to keep the variable point $\left(x,y\right)$(x,y) the same distance away from the given fixed points.
Using the distance formula, this means setting:
$\sqrt{\left(x-5\right)^2+\left(y-3\right)^2}=\sqrt{\left(x-2\right)^2+\left(y-8\right)^2}$√(x−5)2+(y−3)2=√(x−2)2+(y−8)2 and simplifying:
| $\sqrt{\left(x-5\right)^2+\left(y-3\right)^2}$√(x−5)2+(y−3)2 | $=$= | $\sqrt{\left(x-2\right)^2+\left(y-8\right)^2}$√(x−2)2+(y−8)2 |
| $\left(x-5\right)^2+\left(y-3\right)^2$(x−5)2+(y−3)2 | $=$= | $\left(x-2\right)^2+\left(y-8\right)^2$(x−2)2+(y−8)2 |
| $x^2-10x+25+y^2-6y+9$x2−10x+25+y2−6y+9 | $=$= | $x^2-4x+4+y^2-16y+64$x2−4x+4+y2−16y+64 |
| $-10x-6y+34$−10x−6y+34 | $=$= | $-4x-16y+68$−4x−16y+68 |
| $10y$10y | $=$= | $6x+34$6x+34 |
| $y$y | $=$= | $\frac{3}{5}x+3\frac{2}{5}$35x+325 |
Thus the locus is a line with gradient $m=\frac{3}{5}$m=35 and $y$y-intercept $3\frac{2}{5}$325 with the general form given as $3x-5y+17=0$3x−5y+17=0.
More worked examples
Question 1
Question 2
Question 3