The abscissa of the two points A and B are the roots of the equation `x^2+2a x-b^2=0` and their ordinates are the roots of the equation `x^2+2p x-q^2=0.` Find the equation of the circle with AB as diameter. Also, find its radius.

Answer

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

Explanation

Let`(x_(1), y_(1)) and (x_(2), y_(2)) ` be the coordinates of points A and B, respectively.
It is given that `x_(1)+x_(2) " are the roots of " x^(2)+2ax-b^(2)=0`
`rArr" " x_(1)+x_(2)=-2a and x_(1)x_(2)=-b^(2)" "...(i)`
Also, `y_(1) and y_(2) " are the roots of "y^(2)+2py-q^(2)=0`
`rArr y_(1)+y_(2)=-2pand y_(1)y_(2)=-q^(2)" "...(ii)`
`therefore` The equation of circle with AB as diameter is,
`(x-x_(1))(x-x_(2))+(y-y_(1))(y-y_(2))=0`
`rArrx^(2)+y^(2)-(x_(1)+x_(2))x-(y_(1)+y_(2))y+(x_(1)x_(2)+y_(1
)y_(2))=0`
`rArr x^(2) + y^(2)+2ax+2py-(b^(2)+q^(2))=0`
and radius `=sqrt(a^(2)+p^(2)+b^(2)+q^(2))`