InterviewSolution
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InterviewSolution
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Two circles with centres O and O' of radii 3 cm and 4 cm, respectively intersect at two points P and Q such that OP and O'P are tangents to the two circles. Find the length of common chord PQ. |
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Answer» Solution :PO' is a tangent on circle `C_(1)` at P. OP is tangent on circle `C_(2)` at P. As radius OP and tangent PO' are at a point of contact P. `:.""angleP=90^(@)` So, by Pytagoras theorem in right angled `triangleOPO',` `OO'^(2)=OP^(2)+PQ^(2)=3^(2)+4^(2)=9+16=25 cm` `implies""OO' = 5 cm` `triangleOO'P~=triangleOO'Q""`(by SSS criterion of congruence) `implies""angle1=angle2""`(c.p.c.t.) `triangleON'P~=triangleON'Q""`(by SAS criterion of congruence) `implies""angle3=angleO'NQ""`(cpct) `implies""angle3=angleO'NQ=90^(@)""`(LINEAR pair axiom) Let `ON=y`, then `NO'=(5-y)` Let PN = x By Pythagoras theorem in `trianglePNO` and `trianglePNO', we have `x^(2)+""y^(2)""=3^(2)""...(1)`{:(,,x^(2),+,y^(2),=3^(2)"........"(i)),(,,x^(2),+,(5-y)^(2),=4^(2)),(x^(2),+,25,+,y^(2)-10y,=16"........"(2)),(,,x^(2),+,y^(2),=9 ["from" (1)]),(,-,,-,,""-):}/` `25-10y=7""`["SUBTRACT(1) from (2)"] `implies""-10y=7-25""implies""-10y=-18` `implies""y=1.8` But `""x^(2)+y^(2)=3^(2)""`["from (1)"] `implies""x^(2)+(1.8)^(2)=3^(2)""implies""x^(2)=9-3.24` `implies""x^(2)=5.76""implies""x=2.4` `:.` The perpendicular drawn from the centre bisects the chord. `:.""PQ=2PN=2x=2xx2.4=4.8 cm` 
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