# Locus of solutions to an equation?

Describe in detail the locus of points in the x-y plane satisfying this equation:

(x² + y² - 14x + 8y + 65)(7x² + 16xy - 15y² - 7x + 5y) = 0

### 3 Answers

- ?Lv 710 years agoFavourite answer
The first factor appears to be a circle, but completing the square yields:

(x - 7)^2 + (y + 4)^2 = 0

The radius is 0, so it is just a point, namely (7, -4).

The second factor can be factored further to:

(7x - 5y)(x + 3y - 1)

Which describes two lines:

7x - 5y = 0 and x + 3y - 1 = 0

-5y = -7x and 3y = -x + 1

y = 7x/5 and y = -x/3 + 1/3

So the locus is the point (7, -4) and the lines y = 7x/5 and y = -x/3 + 1/3.

- Ian HLv 710 years ago
[(x -7)^2 + (y +4)^2]*[(x + 3y -1)(7x - 5y]= 0

first expression is like circle with zero radius

Is it two intersecting lines ?

3y = 1 - x

5y = 7x

Not too sure

Wolfram might clarify

http://www.wolframalpha.com/examples/DifferentialE...

Regards - Ian

- DIGIMANLv 710 years ago
I got two lines that cross near zero. One line goes thru zero with a positive slope and the other is a

line with negative slope not thru zero but close.

And then there's a spherical object at (7,-4)

Source(s): Dreams and loss of sleep