The intercept on line `y = x` by circle `x^2 + y^2- 2x = 0` is AB. Find equation of circle with AB as a diameter.

Options

A. `x^(2) + y^(2) + x+y = 0`
B. `x^(2) + y^(2) - x - y = 0`
C. `x^(2) + y^(2) - 3x + y = 0 `
D. `x^(2) + y^(2) + 3x - y = 0`

Answer

B

Explanation

We have equation of line y = x and equation of circle `x^(2) + y^(2) - 2x = 0`

Now, interecting points of given line and circle,
`x^(2) + x^(2) - 2x = 0`
`implies 2x^(2) - 2x = 0`
`implies 2x(x - 1) = 0`
`implies x = 0, 1`
When x = 0 then y = 0 and when x = 1 then y = 1
`:.` Coordinates or end points of diameter AB are (0,0) and (1,1)
`:.` Required equation of circle with diameter AB
`(x - 0)(x - 1 + (y - 0) (y - 1) = 0`
`implies x^(2) - 2 + y^(2) - y = 0`
`implies x^(2) + y^(2) - x - y = 0`