Smallest circle that cuts two other circles orthogonally

analytic geometrycirclesconic sectionscoordinate systems

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

Best Answer

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?

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