Question

Two circles with centre o and o' touch of x. Oo' produced meets the circle with centre o' at a

Answer 1

we know that ∠ADO' = 90° ( since O'D is perpendicular to AC)

∠ACO= 90° ( OC(radius)perpendicular to AC(tangent))

In triangles ADO'and ACO ,

∠ADO' = ∠ACO ( each 90°)

∠DAO = ∠CAO (common)

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

AO'/AO = DO'/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 )

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

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

DO'/CO =1/3.

Here's your answer !!

But please write full question buddy!!!