# Intersection of two circles dividing lune in given ratio

If have two circles intersecting along their diameter, let $A,B$ with radius $r_1, r_2$ in the ratio of $1:n/2$ (for $n$ being a positive real number), and overlapping area is the same as radius of smaller circle (let, $r_1$); then how to prove geometrically that the lune formed will be having the intersection point (shown as point $H$, on the line joining the diameters) with length from the center of the smaller circle as : $\frac{r_1}{n}$. I can prove for different ratios of two circles individually and trying algebraically the two equations of different circles with diameters lying only on the $x$-axis.
Say, for ratio $n=5$, the two equations are given below with $b =r_1+2\cdot r_2$ being the total length of the two circles with intersecting length equal to $r_1$, assuming $r_1= \frac{b}{2^3}, r_2 = \frac{5r_1}{2}$ (or, diameter of circle $B=5r_1$) :
$$circle \,\,A :> (x-\frac{b}{8})^2 + y^2 = (\frac{b}{8})^2$$ $$circle \,\,B :> (x-\frac{7b}{16})^2 + y^2 = (\frac{5b}{16})^2$$

Have drawn at desmos too, the graph for the above at : https://www.desmos.com/calculator/e6dhp99eiu.
The graph also shows that the intersection point (shown by dotted line) is having length = $\frac{r_1}{5}$.

But a geometrical proof is needed.

• Please point out any error in problem construction. Apr 17 '18 at 6:22

Using Andrei's notation: Since $C_AK$ is a diameter, $$\angle C_AMK=90^\circ.$$ Hence $$\angle C_AMK = \angle C_AHM.$$

Also $\angle MC_AK = \angle HC_AM.$

Hence $\triangle C_AMK$ is similar to $\triangle C_AHM$, hence we have

$$\frac{C_AH}{C_AM}=\frac{C_AM}{C_AK}=\frac{r_1}{2r_2}=\frac{1}{n}$$

$$\frac{C_AH}{C_AM}=\frac{1}{n}$$

$$\frac{C_AH}{r_1}=\frac{1}{n}$$

$$C_AH=\frac{r_1}{n}$$

• Want to draw in desmos the graph of 'Treasure hunt' problem, with varying position of gallows; but am unable to use the intersecting circles approach that has pine and oak as centers of two different circles, or alternate configuration that places pine and oak on x-axis, for different reasons. Also, have put a post at : math.stackexchange.com/q/2744464/424260 Apr 19 '18 at 11:36
• Please help with the above comment, as the general cases are represented by the $2$nd & $3$rd diagrams, for which getting no idea. Apr 20 '18 at 1:19
• will look at it when i m freer. I thought someone answered that... Apr 20 '18 at 1:23
• Please see the post and my comments. Sorry, for not waiting. But am just lost and in total mess. So, will attempt the problem after your response. Apr 20 '18 at 8:57
• i will take at least $24$ hours and even up to one week... but yup, let's see how things go.. haven't read what is needed exactly yet. Apr 20 '18 at 9:12

Let's call the upper intersection point $M$, and the two centers $C_A$ and $C_B$. The triangles $C_AMH$ and $C_BMH$ are right angle triangles, so we can write $MH$ from Pythagoras' theorem in each triangle: $$MH^2=C_AM^2-C_AH^2=C_BM^2-C_BH^2=C_BM^2-(C_AC_B-C_AH)^2$$ We notice that $C_BM=C_AC_B$, so from the second and fourth terms above $$C_AM^2-C_AH^2=C_AC_B^2-C_AC_B^2+2C_AC_B\cdot C_AH-C_AH^2\\ C_AM^2=2C_AC_B\cdot C_AH\\C_AH=\frac{C_AM^2}{2C_AC_B}=C_AM\frac{1}{2n/2}=\frac{r_1}{n}$$

• Thanks a lot. It is the correct framing of the problem that mattered, particularly the ratio ($r_1: r_2 = 1: n/2$), apart from the appln. of Pythagoras' theorem. Apr 17 '18 at 6:44