A circle is passing through the points A (1, 1) and B (1, 3) and the bisector of first and third quadrant is normal to it, then its area is

Options

A. `2pi`
B. `4pi`
C. `(pi)/(2)`
D. none of these

Answer

A

Explanation

It is given that y=x is a normal to the circle. So, its centre lies on y=x.
C (t, t) be the coordinates of the centre.
AB is a chord of the circle. So, the perpendicular bisector of AB passes through the centre of the circle.
The equation of the perpendicular bisector of AB is y = 2.
Thus, the coordinates of the centre are (2, 2).
`:.` Radius = CA=`sqrt((2-1)^(2)+(2-1)^(2))=sqrt(2)`.
`:.` Area of the circle `= 2 pi` sq. units.