Simultaneous Quadratic Equations (Coordinate Geometry)

[Guest]

Coordinates of a point P3, given Point P1 & P2 at distance A & B respectively:

There are two points P1 at (X1,Y1) and P2 at (X2,Y2).

What are the coordinates of the two points, P31 & P32, which are at a distance A from point (X1,Y1) and a distance B from point (X2,Y2), given that (A+B) > the distance between P1 and P2 which is sqrt((X2-X1)^2 + (Y2-Y1)^2)?

The two solutions for (x,y), as the coordinates of P31 & P32, would satisfy:

(x-X1)^2+(y-Y1)^2=A^2
(x-X2)^2+(y-Y2)^2=B^2

How are these two simultaneous quadratic equations solved for (x,y)? Or, is there an easier way?

[Guest]

You can solve the simultaneous equations, you will eventually get a quadratic equation in one variable (because there are two solutions).

Alternatively you can use trigonometry, apply the cosine rule to work out the angles in the triangle given by P1, P2, P3, and work out the slope between P1 and P2 to get the information you need to find P3.

Either way the answer is pretty horrific, your best bet is to use a math package like mathematica, or matlab, etc. to solve the simultaneous equations otherwise there is a high chance you'll make an error when manipulating the equations.

I'll give an outline:
Expand and subtract the two equations to get
[tex]2(X_2-X_1)x+X_1^2-X_2^2+ 2(Y_2-Y_1)y+Y_1^2-Y_2^2 = A^2-B^2[/tex]
This is a linear equation in [tex]x[/tex] and [tex]y[/tex] and if P1 and P2 are not the same point we can either write [tex]x[/tex] in terms of [tex]y[/tex] or [tex]y[/tex] in terms of [tex]x[/tex] (or both). Substitute this back into one of the quadratic equations and expand to get a quadratic equation in one variable which in theory we can solve (but it will be a mess).

There is no non-messy way of doing it, as the answer is necessarily a complicated expression.

Hope this helped,

R. Baber.

[Guest]

See http://math.stackexchange.com/questions ... -intersect
Salix alba gives a particularly neat solution, by first doing a change of coordinates then rotating back.

Hope this helped,

R. Baber.