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Question
In Fig. 7, two equal circles, with centres O and O’, touch each other at X. OO’ produced meets the circle with centre O’ at A. AC is tangent to the circle with centre O, at the point C. O’D is perpendicular to AC. Find the value of `(DO')/(CO')`
Solution
AO’ = O’X = XO = OC …..(Since the two circles are equal.)
So, OA = AO’ + O’X + XO …..(A-O’-X-O)
∴ OA = 3O’A
In ΔAO'D and ΔAOC,
∠DAO'= ∠CAO ....(Common angle)
∠ADO'= ∠ACO ....(both measure 90°)
∴ ΔADO' ~ ΔACO ....(By AA test of similarity)
`:.(DO')/(CO')=(O'A)/(OA)=(O'A)/(3O'A)=1/3`
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