Two circles of radius $~12~$ and $~3~$ touch externally. A line intersecting both of them intersects first circle at points $P$ and $Q$, second circle - at points $R$ and $S$. Three resulting line segments, two inside the circles and the one between them, are equal: $PQ=QR=RS$. Find their common length.

I have prepared a picture with Geogebra to illustrate

enter image description here

I was trying to solve this with no luck.

After formulating it as a system of equations based on coordinates, with the origin being circles' common point and $X$-axis on the line connecting their centers, Wolfram Alpha helped me to find that the answer should be $\frac{3}{2}\sqrt{13}~$.

Can you give any hints on how to solve this?

  • 3
    $\begingroup$ Lovely problem! $\endgroup$ Commented Jul 20, 2019 at 4:08

2 Answers 2


Let $PQ=QR=RS=2x$, $AM$ and $O_2N$ be perpendiculars to $PS$

and $O_2K$ be a perpendicular to $AM$.

Thus, since $$KO_2=MN=x+2x+x=4x,$$ $$AK=\sqrt{12^2-x^2}-\sqrt{3^2-x^2}$$ and $$AO_2=12+3=15,$$ by the Pythagoras's theorem for $\Delta AO_2K$ we obtain: $$(4x)^2+\left(\sqrt{12^2-x^2}-\sqrt{3^2-x^2}\right)^2=15^2.$$ Can you end it now?

I got $PQ=QR=RS=\frac{3\sqrt{13}}{2}.$

  • $\begingroup$ On which triangle did you apply Pythagoras's Theorem? $\endgroup$
    – eem
    Commented Jul 20, 2019 at 5:28
  • $\begingroup$ @Aditya Dutt I added something. See now. $\endgroup$ Commented Jul 20, 2019 at 5:31
  • $\begingroup$ $PQ=QR=RS=6$ is not a plausible answer because then in $\triangle O_2NR$ you have $NR=3=\frac{1}{2}RS$ and $O_2R=3$ as well. So hypotenuse is same as base length, which is definitely not possible. I think your error is you have put $AK=9$ which is NOT correct. $\endgroup$
    – Anurag A
    Commented Jul 20, 2019 at 6:03
  • 1
    $\begingroup$ @Anurag A Thank you! I fixed. But the idea is the same. $\endgroup$ Commented Jul 20, 2019 at 6:16

Let $PQ=QR=RS=2\ell$.

Drop a perpendicular from center $A$ onto chord $PQ$, let that point be $J$. Similarly drop a perpendicular from the center $O_2$ onto chord $RS$ and call it $K$.

Consider the right $\triangle AJP$, we have $$AJ=\sqrt{144-\ell^2}.$$ Likewise in right $\triangle O_2KR$, we have $$O_2K=\sqrt{9-\ell^2}.$$ Now draw the line $O_2T$ which is parallel to $JK$ (same as saying parallel to the line intersecting the two circles). Then $O_2KJT$ forms a rectangle. Now $$AT=AJ-TJ=AJ-O_2K=\sqrt{144-\ell^2}-\sqrt{9-\ell^2}.$$ Consider the right $\triangle ATO_2$. Observe that $AO_2=12+3=15$ and $O_2T=JQ+QR+RK=4\ell$. So $$AT^2+O_2T^2=AO_2^2 \implies \left[\sqrt{144-\ell^2}-\sqrt{9-\ell^2}\right]^2+(4\ell)^2=15^2.$$ So we need to solve for $\ell$.

Once you simplify this equation by appropriately squaring etc. you end up getting $$\ell^2(48\ell^2-351)=0.$$ This implies $\ell^2=\frac{117}{16}$. Consequently, $$\color{red}{2\ell=\frac{3\sqrt{13}}{2}}.$$


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