Find the locus of a point...

Question

The locus of a point is the path made by that point when it moves so as to satisfy a given geometric condition. In recent times the word locus has sometimes been replaced by the term "set of points. Example: the locus of a point which is equidistant from 2 fixed point is the perpendicular bisector of the line segment joining the points.

Find the locus of a point such that its distances from 2 given points are in the ratio of 3:4.

Answer 1

Wow, what an intriguing question! I couldn't resist figuring it out. What I did was just picked two points (a,b) and (c,d) and created an equation for a set of points (x,y) where the ratio of the distance from (x,y) to (a,b) and the distance from (x,y) to (c,d) is 3/4:

√((x-a)²+(y-b)²) / √((x-c)²+(y-d)²) = 3/4

Then I squared both sides to get rid of the radicals:

((x-a)²+(y-b)²) / ((x-c)²+(y-d)²) = 9/16

And cross-multiplied:

16(x-a)² + 16(y-b)²) = 9(x-c)² + 9(y-d)²

You could expand it out further, but even without doing so, you'll notice that the equation contains x and y expressions to the second power with the same coefficient. That's an equation of a circle! So the answer is that this locus forms a set of points that form a circle!

UPDATE: I've been trying to see if I can come up with something more specific, such as being able to identify the center of the circle in relation to the two points and the 3/4 ratio, but I'm coming up empty. If anyone can figure it out, I'd love to see it!

UPDATE 2: I've worked it out by multiplying it out and completing the squares, but it's not pretty. The locus is a circle centered at ( (16a-9c)/7, (16b-9d)/7 ) with a radius of √( (9c²+9d²-16a²-16b²)/7 + (16a-9c)²/49 + (16b-9d)²/49 ), where (a,b) and (c,d) are the two given points in the question. I was hoping for something more elegant at the end.