The equation of chord AB of the circle `x^2+y^2=r^2` passing through the point P(1,1) such that `(PB)/(PA)=(sqrt2+r)/(sqrt2-r)`, `(0ltrltsqrt(2))`

Answer

`y=x`

Explanation

Chord through P(1,1) meets the circles at A and B.
`:. PA xx Pb=` ( length of tangent `)^(2) = PT ^(2)`
or `PA xx PB =2 -r^(2)` (1)
Given `(PB)/(PA)=(sqrt(2)+r)/(sqrt(2)-r)` (2)
Solving (1) and (2), we get
`PB =sqrt(2)+r,PA=sqrt(2)-r`
`implies AB=PB-PA=2r=` diameter of circle
Therefore, equation of chord AB is `y=x`