Two circles and the max length of a line through their intersection

Question

Consider two circles and their intersections A and B. Let r be a line through A, and call C and D the other intersections between r and the two circles. Prove that CD is of maximum length iff r is parallel to the line that passes through the center of both circles. No calculus allowed (this is a tenth grade problem).

My initial approach was to draw r and r', one parallel one not, and try to use the triangular inequality. This doesn't work because while one segment (say, AD) decreases the other (AC) increases.

Trying to use algebraic formulas to compute the lengths as a function of the radius of both circles gets messy, and given the context (this is a problem from a tenth grade textbook) I doubt this is necessary. There's bound to be a simple geometric solution I'm missing.

I tried to see if an additional assumption of equal radius for both circles simplified things or led to some insight but it didn't look like it. I tried to see if considering the triangles formed by r with the line through the centers could help, but it felt far off.

The thing that felt the closest to a solution was the idea of considering the chords AC and AD as being tied to the angles on the circumference, and somehow find angles that geometrically sum to something (either adjacent or part of some triangle). There's still the problem of having a different radius on the two circles, and even then, how would that turn into a condition on r being parallel?

I'm pretty stumped here.

Answer 1

Consider arbitrary line IJ. Drop perpendiculars from O and O', they meet IJ at G and H. Trapezoid OGHO' is right angled at H , so $OO'>GH=\frac{IG}{2}$. you want IJ be maximum, This means trapezoid must become rectangle such that $GH=OO'=EF=\frac{CD}2$, that is G and H become coincident on F and E resprctively. where quadrilateral OEFO' is rectangle and $OO'=FE$, that is CD must be parallel with OO'.

Answer 2

A different approach using a little calculus and trigonometry. The triangle $\triangle{M_1M_2A}$ is fixed. Line DF rotates around A over an angle $\varphi$ which is the only variable. Length DF can be easily expressed in terms of $\varphi$. To get an optimum, we differentiate and establish that $\frac{sin(\varphi+\alpha)}{r_2}=\frac{sin(\varphi)}{r_1}$ Now apply the sine rule in $\triangle{M_1 M_2 A}$. This results in $\frac{sin(\epsilon+\alpha)}{r_2}=\frac{sin(\epsilon)}{r_1}$ Therefore the maximum length DF is reached when $\epsilon=\varphi$ Which implies that DF is parallel to the line passing through the center of both circles.