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

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?

## marked as duplicate by Carl Mummert, Community♦May 19 '15 at 14:22

• There is a nice article in wolfram: mathworld.wolfram.com/Circle-CircleIntersection.html – SamC May 19 '15 at 13:41
• This isn't a duplicate of the other question of which it's marked a duplicate (at least, the way all answers to the other question seem to be interpreting it-- the other question isn't completely clear). This is a simpler question, because the two radii are equal. – Don Hatch Jul 4 '18 at 0:14

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...

• Shouldn't there be a factor of $2$ in the denominator? – Demosthene May 19 '15 at 13:38
• Should be $y=\dfrac{(a_2-a_1)(2x-a_1-a_2)}{2(b_1-b_2)}+{(b_2+b_1) \over 2}$ – SamC May 19 '15 at 13:39
• Just an aside, I suggest leave the answer as this form $(a_2-a_1)(2x-a_1-a_2)=(b_1-b_2)(2y-b_2-b_1)$, since $b_1$ and $b_2$ can be $0$ as the same time. i.e. Two circles that have center $(0,0)$ and $(2,0)$ with radii $1$. – SamC May 19 '15 at 13:47

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.

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.