# Equal intersection segments on circles

On a plane, there are two congruent circles, $$O_1, O_2$$. One line intersects $$O _1$$ and A and B and $$O_2$$ at C and D such that AB=BC=CD=14. Another line intersects $$O _1$$ and E and F and $$O_2$$ at G and H such that EF=FG=GH=6. What is the radius of the circle.

The configuration was hard to draw. https://www.geogebra.org/geometry/tgxp8dqa. You can reflect the lines over the centers, but it is still the same. Cleary, $$\widehat{AB}=\widehat{CD}$$ and $$\widehat{EF}=\widehat{GH}$$. I have no idea how to proceed.

• I do not see the points E, F,G in the figure. Also, should we assume that the centers are on one side of one of the lines and on opposite sides of the other one? – GReyes Oct 9 at 15:15

\begin{align} |AB|&=|BC|=|CD|=u=14 ,\\ |EF|&=|FG|=|GH|=v=6 . \end{align}
\begin{align} |O_1O_2|&=2u , \end{align}
\begin{align} \triangle O_1IN:\quad& \\ |O_1I|&=\tfrac12\,|O_1O_2|=u ,\\ |NI|&=\tfrac12\,|EF|+\tfrac12\,|FG|=v ,\\ |O_1N|^2&=u^2-v^2 \tag{1}\label{1} , \end{align}
\begin{align} \triangle O_1NE:\quad& \\ |O_1E|&=R ,\\ |NE|&=\tfrac12\,|EF|=\tfrac v2 ,\\ |O_1N|^2&=R^2-(\tfrac v2)^2 \tag{2}\label{2} . \end{align}
It follows from \eqref{1}$$=$$\eqref{2} that
\begin{align} R^2&=u^2-\tfrac 34\,v^2 ,\\ R&=13 . \end{align}