Q17 of 31 Page 65

The two circles with their centres at P and Q intersect each other at the points A and B. Through the point A, a straight line parallel to PQ intersects the two circles at the points C and D respectively. If PQ = 5 cm., then let us determine the length of CD.


Given: PQ = 5cm, CD is parallel to PQ, Two Circles with centers P and Q intersect at point A and B.


Construction: EF and FQ are Perpendicular bisector drawn from P and Q Respectively.



The Perpendicular from The Centre to the chord, bisects the chord.


⇒ DE = EA


⇒ CF = AF


⇒ DE + EA + AF + FC = DC


⇒ 2EA + 2AF = DC


⇒ 2(EA + AF) = DC ………..(1)


EFPQ (Given)


PEA = 90o (Construction)


QFA = 90o (Construction)


Using Interior Angle Theorem,


PEA + BPE = 180


BPE = 90o


PQFE is a rectangle.


So, EF = PQ …………… (2)


⇒ 2(EA + AF) = DC


⇒ 2EF = DC


⇒ 2PQ = CD


CD = 10cm


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