Given two circles $(x_1, y_1, r_1), (x_2, y_2, r_2)$ and a line passing through two points $A(x_a, y_a)$ and $B(x_b, y_b)$. How to find a circle $(x_3, y_3, r_3)$ that is tangent to line and two given circles?

I need an algebraic equations not geometric construction.


Let's say $\epsilon$ is the line defined by A and B and $K_1,K_2$ and $K_3$ the centers of the given circles.

Then we have the following 3 equations that will help us define $x_3,y_3,r_3$

$1$. $\sqrt{(x_3-x_2)^2+(y_3-y_2)^2}=r_3+r_2$ or $\sqrt{(x_3-x_2)^2+(y_3-y_2)^2}=|r_3-r_2|$

That is, the distance of the centers $K_2,K_3$ equals the sum of the radius $r_3,r_2$ if the circles are tangent outwardly or the $|r_3-r_2|$ if they're tangent inwardly.

$2$. $\sqrt{(x_3-x_1)^2+(y_3-y_1)^2}=r_3+r_1$ or $\sqrt{(x_3-x_1)^2+(y_3-y_1)^2}=|r_3-r_1|$

$3$. The distance of the $K_3$ from $\epsilon$ equals $r_3$ if you write $\epsilon :Ax+By+C=0$ then $r_3=\frac{|Ax_3+By_3+c|}{\sqrt{A^2+B^2}}$

  • $\begingroup$ We obtain tree equation (1)-(3) where x3, y3, r3 are unknown. How to solve them? $\endgroup$ – Ivan Bunin Jan 12 '13 at 21:42
  • $\begingroup$ From equation 3 you can easily compute $r_3$ just substitute everything you know. Then substitute everything you know in equations 1 and 2 and subtract equation 1 from 2 (or 2 from 1 it's the same) to get rid of the squares. Solve what is left so that $x_3=....$ substitute again and you're done... Lot's of things to do I know... $\endgroup$ – epsilon Jan 12 '13 at 22:00
  • $\begingroup$ I've solved the problem using Wolfram MathWorld page and some additional calculation. Thank you! $\endgroup$ – Ivan Bunin Jan 13 '13 at 12:37
  • $\begingroup$ Glad I could help :) $\endgroup$ – epsilon Jan 13 '13 at 13:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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