CIrcle C (r=1) touches X axis at A.Centre Q of C lies in Quad I.Tangent from origin touches C at T and P lies on it such that OAP is rt triangle (at A).perimeter =8
Q1 Find PQ
Q2 eqn of circle
Q3 eqn of tangent
here is fig...plz give hints or soln..
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2 Answers
eragon24 _Retired
·2010-01-26 06:15:03
OA=OT=a
PT=b
PQ=x
QA=1
since perimetr=8
so a+b+x+1+a=8..............i
anle QOA=angle QOT=β
so tan2β=x+1a........ii
tanβ=1/a...........................iii
b=√(x2-1).............iv
triangle PTQ is similar to triangle POA
SO a+bx=a1........v
now solve