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Equal circles with centres $ O $ and $ O^{\prime} $ touch each other at $ X . O O^{\prime} $ produced to meet a circle with centre $ O^{\prime} $, at $ A . A C $ is a tangent to the circle whose centre is O. $ O^{\prime} D $ is perpendicular to $ A C $. Find the value of $ \frac{D O^{\prime}}{C O} $.
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Given:
Equal circles with centres \( O \) and \( O^{\prime} \) touch each other at \( X . O O^{\prime} \) produced to meet a circle with centre \( O^{\prime} \), at \( A . A C \) is a tangent to the circle whose centre is O. \( O^{\prime} D \) is perpendicular to \( A C \).
To do:
We have to find the value of \( \frac{D O^{\prime}}{C O} \).
Solution:
$AC$ is the tangent of the circle with centre $O$.
$O^{\prime}D\ perp\ AC$ is drawn and $OC$ is joined.
$AC$ is tangent and $OC$ is the radius.
$OC\ \perp\ AC$
$O^{\prime}D\ \perp\ AC$
$OC\ \parallel\ O^{\prime}D$
$OA=O^{\prime}A+O^{\prime}X+OX$
$OA=3AO^{\prime}$ ($O^{\prime}A=O^{\prime}X=OX$
In triangles $O^{\prime}AD$ and $OAC$,
$\angle A = \angle A$ (Common angle)
$\angle AO^{\prime}D =\angle AOC$ (Corresponding angles)
Therefore,
$\Delta \mathrm{O}^{\prime} \mathrm{AD} \sim \Delta \mathrm{OAC}$ (By AA axiom) $\frac{\mathrm{DO}^{\prime}}{\mathrm{CO}}=\frac{\mathrm{AO}^{\prime}}{\mathrm{AO}}$
$=\frac{\frac{1}{3} \mathrm{AO}}{\mathrm{AO}}$
$=\frac{1}{3}$
The value of \( \frac{D O^{\prime}}{C O} \) is $\frac{1}{3}$.
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