A foot of the normal from the point (4, 3) to a circle is (2, 1) and a diameter of the circle has the equation `2x-y-2=0`. Then the equation of the circle is :

Options

A. `x^(2)+y^(2)+2x-1=0`
B. `x^(2)+y^(2)-2x-1=0`
C. `x^(2)+y^(2)-2y-1=0`
D. none of these

Answer

B

Explanation

The equation of the normal at P(2,1) is
`y-1=(3-1)/(4-2)(x-2)rArr x-y-1=0`
Clearly, point C i.e. the centre of the circle is the point of intersection of the lines x-y-1=0 and 2x-y-2=0. So, its coordinates are (1, 0).

Also, radius = CP=`sqrt((2-1)^(2)+(1-0)^(2))=sqrt(2)`
Thus, the equation of the circle is
`(x-1)^(2)+(y-0)^(2)=(sqrt(2))^(2) or, x^(2)+y^(2)-2x-1=0`