Question

Two equal circles with centers O and O' touch each other at X. OO' produced meets the circle with center O' at A. AC is the tangent to the circle with center O, at the point C. O'D is perpendicular to AC. Find the value of DO'/CO.

Answer 1

angleADO'=90 ( since O'D is perpendicular to AC) angleACO= 90 ( OC(radius)perpendicular to AC(tangent))

In triangles ADO'and ACO ,

angleADO'=angleACO ( each 90)

angle DAO = angle CAO (common)

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

AO'/AO=DO'/CO ( corresponding parts of similar triangles )

AO= AO'+O'X+OX

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

AO'/AO=AO'/3AO'=1/3

DO'/CO=AO'AO=1/3

DO'/CO=1/3