The circle described on the line joining the points `(0,1), (a, b)` as diameter cuts the x-axis in points whose abscissae are roots of the equation

Options

A. `x^(2)+ax+b=0`
B. `x^(2)-ax+b=0`
C. `x^(2)+ax-b=0`
D. `x^(2)-ax-b=0`

Answer

B

Explanation

The equation of the circle described on the line joining (0, 1), (a, b) as diameter is given by
`(x-0)(x-a)+(y-1)(y-b)=0`
`rArr x^(2)+y^(2)-ax-y(1+b)+b=0`
This meets x-axis at y=0. Therefore, the abscissae of the points where the circle meets x-axis are roots of the equation `x^(2)-ax+b=0`.