Find the equation of the circle which is touched by `y=x` , has its center on the positive direction of the x=axis and cuts off a chord of length 2 units along the line `sqrt(3)y-x=0`

Options

A. `x^(2)+y^(2)-4x+2=0`
B. `x^(2)+y^(2)-4x+1=0`
C. `x^(2)+y^(2)-8x+8=0`
D. `x^(2)+y^(2)-4y+2=0`

Answer

A

Explanation

Since the required circle has its centre on x-axis. So, let the coordinates of the centre be (a, 0). The circle touches y=x. Therefore,
Radius = Length of the `_|_`
from (a, 0) on x-y=0
`rArr ` Radius `= (a)/(sqrt(2))`
The circle cuts off a chord of length 2 units along `x-sqrt(3)y=0`
`:. ((a)/(sqrt(2)))^(2)=1^(2)+((a-sqrt(3)xx0)/(sqrt(1^(2)+(sqrt(3))^(2))))^(2) rArr (a^(2))/(2)=1+(a^(2))/(4)rArr a=2`
Thus, centre of the circle is at (2, 0) and radius `=(a)/(sqrt(2))=sqrt(2)`
Hence, its equation is `x^(2)+y^(2)-4x+2=0`