2021-12-28

Two circles with a common chord and center of one circle lie on another circle

This question was uploaded on 28/12/21 on social media accounts.

Solution:

Chord with length m+n is a common chord.
Intersecting chord theorem in the purple circle:
\[\color{red} {mn = (r+x)(r-x)}\]
\[\Rightarrow \color{red} {mn = r^2-x^2}\]
Intersecting chord theorem in the green circle:
\[\color{blue} {mn = (x)(2r-x)}\]
\[\Rightarrow \color{blue} {mn = 2rx-x^2}\]

From the above equations:
\[\color{red} {r^2-x^2} = \color{blue} {2rx-x^2}\]
\[\Rightarrow \color{green} {r = 2x}\]

Intersecting chord theorem in the green circle at perpendicular:
\[\color{purple} {(r)(r) = (2R-x)(x)}\]
\[\Rightarrow \color{purple} {(\color{green} {2x})(\color{green} {2x}) = (2R-x)(x)}\]
\[\Rightarrow \color{purple} {R = \frac52x}\]

Now we can calculate the ratio:
\[\frac rR = \frac{\color{green} {2x}}{\color{purple} {(5/2)x}} = \frac45\]
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