Q24 of 47 Page 10

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 .

Given: AC is tangent to the circle with centre O.


To find: the value of .


Theorem Used:


Tangent to a circle at a point is perpendicular to the radius through the point of contact.


Explanation:


AC is the tangent to radius O'D.


So, by the theorem stated O'D ⊥ AC


⇒ ∠ADO' = 90°


Similarly, ∠ACO= 90°


In ΔADO' and ΔACO,


∠ADO' = ∠ACO (each 90°)


∠DAO = ∠CAO (common)


by AA criterion.


ADO' ~ ACO


(corresponding sides of similar triangles)


AO = AO' + O'X + OX


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





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