Q11 of 31 Page 65

The centers of two circles are P and Q; they intersect at the points A and B. The straight line parallel to the line-segment PQ through the point A intersects the two circles at the points C and D. I prove that, CD = 2PQ.


Given, CD is parallel to PQ, Two Circles with centers P and Q intersect at point A and B.


To Prove: 2PQ = CD


Construction: PE and QE are the Perpendiculars on the Chord CD From centers P and Q respectively.


⇒ DE = EA


⇒ CF = AF


⇒ DE + EA + AF + FC = DC


⇒ 2EA + 2AF = DC


⇒ 2(EA + AF) = DC ……..(1)


PQ and CD are parallel lines and PE and QE are the perpendiculars.


So, AEP = 900,AFQ = 900,EPQ = 900,AFQ = 900


EPQF is a rectangle


So, EF = PQ ……… (2)


⇒ 2(EA + AF) = DC


⇒ 2EF = DC


⇒ 2PQ = DC


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