How do you find the intersection(s) of two circles with equal radii?

Question

I have two circles with the following equations: \begin{equation*} (x-a_1)^2+(y-b_1)^2=r^2 \\ (x-a_2)^2+(y-b_2)^2=r^2 \end{equation*} The two radii are equal. How do you find the intersections of any two circles with equal radii?

Answer 1

Notice that you have $(x-a_1)^2+(y-b_1)^2=r^2=(x-a_2)^2+(y-b_2)^2$

That you transform in

$(x-a_1)^2-(x-a_2)^2=(y-b_2)^2-(y-b_1)^2$

$a^2-b^2=(a-b)(a+b)$ thus you have

$(a_2-a_1)(2x-a_1-a_2)=(b_1-b_2)(2y-b_2-b_1)$

or

EDIT: forgot the factor 2, thanks for the comments!

$y=\dfrac{(a_2-a_1)(2x-a_1-a_2)}{2(b_1-b_2)}+\dfrac{b_2+b_1}{2}$

Don't forget that you have also to verify the first equations, so you have to plug this into one of them...

Answer 2

Simply equate them: the points of intersection belong to both circles, and thus verify $$(x_i-a_1)^2+(y_i-b_1)^2=(x_i-a_2)^2+(y_i-b_2)^2$$ where $(x_i,y_i)$ are the coordinates of the intersection points.

Answer 3

Simplify the equations, subtract one from,the other, express y trough x or x through y, and then put it in one of the original equations.

Now you can find the intersection points.