a circle passing through the point `(2,2(sqrt2-1))` touches the pair of lines `x^2 -y^2 -4x+4 =0`. The centre of the circle is

Options

A. `(2,2sqrt(2))` and `(2,6sqrt(6)-8)`
B. `(2,5sqrt(2))` and `(2,7sqrt(2))`
C. `(2,5sqrt(2)-1)` and `(2,-3)`
D. None of these

Answer

A

Explanation

The equation of the pair of lines is
`rArr (x-2)^(2) -y^(2) =0`
`rArr (x-2) = +-y`
i.e. `x -y =2` and `x +y =2`
The centre of the circle touching the above lines must lie on the angular bisectors of the above lines. Hence, `C -= (2,k)`

Thus, we have
`rArr (|2+k-2|)/(sqrt(2)) = 2(sqrt(2)-1)-k`
`rArr +-(k)/(sqrt(2)) =2 (sqrt(2)-1)-k`
`rArr k(1+-(1)/(sqrt(2))) =2 (sqrt(2)-1)`
`rArr k =(2sqrt(2)(sqrt(2)-1))/(sqrt(2)+-1)`
`= 2sqrt(2),2sqrt(2) (sqrt(2)-1)^(2)`
`= 2sqrt(2),6sqrt(2) -8`