-1
$\begingroup$

This question already has an answer here:

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?

$\endgroup$

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ There is a nice article in wolfram: mathworld.wolfram.com/Circle-CircleIntersection.html $\endgroup$ – SamC May 19 '15 at 13:41
  • 1
    $\begingroup$ 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. $\endgroup$ – Don Hatch Jul 4 '18 at 0:14
2
$\begingroup$

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...

$\endgroup$
  • 1
    $\begingroup$ Shouldn't there be a factor of $2$ in the denominator? $\endgroup$ – Demosthene May 19 '15 at 13:38
  • $\begingroup$ Should be $y=\dfrac{(a_2-a_1)(2x-a_1-a_2)}{2(b_1-b_2)}+{(b_2+b_1) \over 2}$ $\endgroup$ – SamC May 19 '15 at 13:39
  • $\begingroup$ 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 $1$. $\endgroup$ – SamC May 19 '15 at 13:47
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.