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In figure, two equal circles, with center $O$ and $O'$, touch each other at X. $OO'$ produced meets the circle with center $O'$ at $A$. $AC$ is tangent to the circle with center $O$, at the point C. $O'D$ is perpendicular to AC. Find the value of $\frac{DO'}{CO}$.
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Given: Two equal circles, with center O and O', touch each other at X. OO' produced meets the circle with center O' at A. AC is tangent to the circle with center O, at the point C. O'D is perpendicular to AC in the given figure.

To do: To Find the value of $\frac{DO'}{CO}$.

Solution:

$AO’=O’X = XO= OC\ \ \ \ \ \ \ \ \ \ \dotsc ..( Since\ the\ two\ circles\ are\ equal.)$

So, $OA=AO’ + O’X + XO\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \dotsc ..( A-O’-X-O)$

$\therefore \ OA\ =\ 3O’A$

In $\vartriangle AO'D$ and $\vartriangle AOC$,

$\angle DAO' = \angle CAO \ \ \ \ \ \ \ \ ....\ ( Common\ angle)$

$\angle ADO'=\angle ACO\ \ \ \ \ \ \ \ ....\ ( both\ measure\ 90^{o} )$

$\vartriangle ADO'\ \sim \vartriangle ACO \ \ \ \ \ ....\ ( By\ AA\ test\ of\ similarity)$

$\frac{DO'}{CO} =\frac{O'A}{OA} =\frac{O'A}{3O'A} =\frac{1}{3}$
 

Updated on: 10-Oct-2022

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