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 tangent to the circle whose centre is O. O’ D is perpendicular to AC. Find the value of \(\frac{DO'}{CO}.\)

Answer 1

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

As we know radius is perpendicular to the tangent.

So, OC ⊥ AC

⇒ ∠ACO = 90° 

In ΔADO’ and ΔACO, 

∠ADO’ = ∠ACO (each 90°) 

∠DAO = ∠CAO (common) 

By AA criteria, 

ΔADO’ ∼ ΔACO

As we know corresponding sides of a triangle are in ratio.

AO = AO’ + O’X + OX

As radii of two circles are equal.

⇒ AO = AO’ + AO’ + AO’

= 3 AO’