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} $."

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 $\frac{1}{3}$.