Question

Equal circles with centres 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

we know that $\angle AD{O}^{\prime }$ = 90° ( since O'D is perpendicular to AC)

$\angle AC{O}^{\prime }$ = 90° ( OC(radius)perpendicular to AC(tangent))

In triangles ADO'and ACO ,

$\angle AD{O}^{\prime }$ = $\angle AC{O}^{\prime }$ ( each 90°)

$\angle DA{O}^{\prime }$ = $\angle CAO$ (common)

by AA criterion ,triangles ADO' and ACO are similar to each other.

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

( corresponding sides of similar triangles )

AO = AO' + O'X + OX

= 3AO' (since AO'=O'X=OX because radii of the two circles are equal )

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

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

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