Question

Find the intersection of 2 circles

Answer 1

Circle with point A(Xa,Ya) with radius m
Circle with point B(Xb,Yb) with radius n
Point P(x,y) is where A and B intersect
find the point(s) of intersection

Answer 2

So basically, we have:
(x-Xa)^2+(y-Ya)^2=m^2
(x-Xb)^2+(y-Yb)^2=n^2
and we have to solve for x and y.
I kinda get lost at the plus/minus bit

Answer 3

Feel free to change pronumerals to whatev suits u

Answer 4

hmm, one way ... is to find the midpoint and slope between centers;

define the line perp to that going thru the midpoint;

then see where that hits either circle

Answer 5

prolly aint the kosher solution; but its bound to get results

Answer 6

Yeah, kinda looking to extending this out to find the intersection between 3 circles (triangulation)

Answer 7

or .....

take waht you have, subtract the m^2 and n^2 from the sides to get them both equal to zero

then equate the 2 and see what simplifies

Answer 8

(x-Xa)^2+ (y-Ya)^2 - m^2
-(x-Xb)^2 - (y-Yb)^2 + n^2
--------------------------
prolly a linear in the end

Answer 9

best to expand out the ^2 parts tho

Answer 10

hmm... ok will try

Answer 11

ewww ...... either i did something horribly wrong; or my first thought was way better lol

Answer 12

is pts A and B centers of the circles? becasue now that i read it again, its a little vague

Answer 13

Yeah, A and B are the centre of circles

Answer 14

@amistre64, back to your first suggestion, how would finding the midpoint help?

Answer 15

\[(\frac{X_a+X_b}{2},\frac{Y_a+Y_b}{2})\]

Answer 16

this gives us an anchor to place the perp slope on

Answer 17

But why the midpoint? Is this significant?

Answer 18

hmm, at first i thought it was ... mighta been a bad notion tho

Answer 19

The midpoint won't necessarily be on the perpendicular though

Answer 20

if the circles aint the same radius; i guess that defeats a midpoint

Answer 21

distance between centers is the base of a triangle; with the radiuses being the other two sides ... might be applicable

Answer 22

essentially you have 2 equations with 2 variables
by graphing the 2 circles you will know if you have 0,1,or 2 solutions
solve this by substitution...solve each equation for y, then set them equal to get x
\[\sqrt{m^{2}-(x-X_a)^{2}} + Y_a = \sqrt{n^{2}-(x-X_b)^{2}}+Y_b\]
from here you can solve for x, there may be no solution
lot of algebra involved