Question

Equal circles with centre O and O' touch each other at X. OO' produced to meet a circle with centre O, at A. AC is a tangent to the circle whose centre is O. O'D is perpendicular to AC. Find the value of $\frac{D{O}^{\prime }}{CO}.$

Answer 1

As the both circles are equal so the radius will be same for both circles

Let the radius of two circles be $r$

$⟹OA=OX+X{O}^{\prime }+O^{\prime }A=r+r+r=3r,O^{\prime }A=r$

Given $O^{\prime }D\perp AC⟹\angle AD{O}^{\prime }={90}^{\circ }$

$AC$ is the tangent which will be perpendicular to the radius $OC⟹\angle ACO={90}^{\circ }$

In $△ACO$ and $△AD{O}^{\prime }$

$\angle A=\angle A$ $(\because \text{common angle})$

$\angle ACO=\angle AD{O}^{\prime }={90}^{\circ }$

By $AA$ similarity

$△ACO\sim △AD{O}^{\prime }$

So $\frac{D{O}^{\prime }}{CO}=\frac{A{O}^{\prime }}{AO}=\frac{r}{3r}=\frac{1}{3}$

Hence the answer is $\frac{1}{3}$