If a line passes through the point `P(1,-2)` and cuts the `x^2+y^2-x-y= 0`at `A` and `B`, then the maximum of `PA+PB` is

Options

A. `sqrt(26)`
B. 8
C. `sqrt(8)`
D. `2 sqrt(8)`

Answer

1

Explanation

Any point on the line passing through the point P(1,-2) and having slope `tan theta` is
`(1+r cos theta ,-2+r sin theta)`
It lies on the given circle.
`:. (1+r cos theta )^(2)-(-2+r sin theta)^(2)`
`- (1+r cos theta)-(-2 +r sin theta)=0`
`:. r^(2)+(cos theta -5 sin theta )r +6=0`
Above equation has roots `r_(1)` and `r_(2)`, where `r_(1)=PA` and `r_(2)=PB`
`:. PA+PB=r_(1)+r_(2)= cos theta -5 cos theta`
`(cos theta-5 sin theta)_(max)=sqrt(26)`