If the abscissa and ordinates of two points `P and Q` are the roots of the equations `x^2+2a x-b^2=0` and `x^2+2p x-q^2=0` , respectively, then find the equation of the circle with `P Q` as diameter.

Options

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

Answer

A

Explanation

Let `x_(1), x_(2)` and `y_(1), y_(2)` be the roots of the equations `x^(2)+2ax-b^(2)=0` and `x^(2)+2px-a^(2)=0`
respectively .
Then,
`x_(1)+x_(2)=-2a, x_(1)x_(2)=-b^(2)`
Then,
`x_(1)+x_(2)=-2a, x_(1)x_(2)=-b^(2)`
and, `y_(1)+y_(2)=-2p, y_(1)y_(2)=-q^(2)`
The equation of the circle with `P(x_(1), y_(1))` and `Q(x_(2), y_(2))` as the end points of diameter, is
`(x-x_(1))(x-x_(2))+(y-y_(1))(y-y_(2))=0`
`rArr x^(2)+y^(2)-x(x_(1)+x_(2))-y(y_(1)+y_(2))+x_(1)x_(2)+y_(1)y_(2)=0`
`rArr x^(2)+y^(2)+2ax+2py-b^(2)-q^(2)=0`.