Question

In the figure, two equal circles, with centres $O$ and $O^{\prime }$, touch each other at $X.O^{\prime }X$ produced meets the circle with centre $O$ at $A.AC$ is tangent to the circle with centre $O,$ at the point $C.$
$O^{\prime }D$ is perpendicular to $AC.$ Find the value of $\frac{D{O}^{\prime }}{CO}$.

Answer 1

$O^{\prime }D$ is perpendicular to $AC.$

We know that $\angle AD{O}^{\prime }={90}^o$

Radius $OC$ is perpendicular to tangent $AC.$

In $\Delta AD{O}^{\prime }$ and $\Delta ACO$,

$\angle AD{O}^{\prime }=\angle ACO$ (each 90 degree)

$\angle DAO=\angle CAO$ (common)

By AA property, trangles ADO' and ACO are similar to each other.

$\frac{A{O}^{\prime }}{AO}=\frac{D{O}^{\prime }}{CO}$ (corresponding sides of similar triangles)

$AO=A{O}^{\prime }+O^{\prime }X+OX$

$=3A{O}^{\prime }$ (Since $A{O}^{\prime }=O^{\prime }X=OX$ because radii of the two circles are equal)

$\frac{A{O}^{\prime }}{AO}=\frac{A{O}^{\prime }}{3AO}=\frac{1}{3}$

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

$\frac{D{O}^{\prime }}{CO}=\frac{1}{3}$.