circles..

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..

2 Answers

21
eragon24 _Retired ·

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

24
eureka123 ·

thx...trying now...[1]

Your Answer

Close [X]