# Smallest circle that cuts two other circles orthogonally?

I need to find the center of the smallest circle which is orthogonal to two other circles. I know that the center of circles orthogonal to two other circles will lie on the radical axis of those two circles.

My book's solution for this question, states that the required center lies on the intersection of the radical axis of the two given points and the line joining their centers.

I do know that it lies on the radical axis, but does the center of the (smallest) circle also lie on the line joining the centers? This is not stated as a fact and isn't particularly obvious to me, so can someone help me with a proof? I tried to prove this but wasn't able to get the conditions for the circle to be "smallest".

I also know that the common chord of two intersecting circles is the same as the radical axis, and the common chord is bisected by the line joining the centers.

Using these, I was not able to come up with a proof, any help would be appreciated greatly

• I believe that once you set up the equations, it all falls out. – Calvin Lin Apr 9 '20 at 15:21
• If the two original circles intersect at two points then their joint chord seems to lie on the radical axis. Can any point on the interior of this chord (including the one on the line joining the two original centres) be the centre of a circle orthogonal to the original circles? – Henry Apr 9 '20 at 15:48
• If the two original circles intersect then (as a limit) the smallest circles are radius $0$ centred at the points of intersection – Henry Apr 9 '20 at 16:19

Note: This deals with the case when the 2 disks do not intersect each other. You will have to deal with the cases when 1) one circle is contained within the other, and 2) one circle intersects another.

Let the 2 circles have centers $$O_1, O_2$$.
Let the radical axis intersect $$O_1O_2$$ at $$A$$.
Let $$B$$ be any point on the radical axis.
Let $$r_B$$ be the radius of the circle that is orthogonal to these 2 circles.

We have $$r_B^2 + r_1^2 = O_1B^2 = OA^2 + AB^2$$.
and $$r_B^2 + r_2^2 = O_2B^2 = OB^2 + AB^2$$.

Hence $$2r_B^2 = (OA^2 +OB^2 - r_1^2 - r_2^2) + 2AB^2$$.
When does the minimum $$r_B$$ occur?

• Could you please elaborate how you came up with those two equations? That $r_B^2+r_1^2=O_1 B^2$ and the other ones? – Techie5879 Apr 9 '20 at 15:31
• Pythagorean theorem on the centers of the circle and one of their intersection points. That's a pretty standard first result of orthogonal circles, which I assumed you knew given that you knew the fact about radical axis being the locus of circles that are orthogonal to both, which can again be derived from this result. – Calvin Lin Apr 9 '20 at 15:33
• Oh thanks, is $r_B$ minimum when $AB$ is minimum? – Techie5879 Apr 9 '20 at 15:34
• What do you think? – Calvin Lin Apr 9 '20 at 15:34
• That statement is correct. – Calvin Lin Apr 9 '20 at 15:36