If the circle `x^2+y^2+2x+3y+1=0` cuts `x^2+y^2+4x+3y+2=0` at `Aa n dB` , then find the equation of the circle on `A B` as diameter.

Answer

`2x^(2)+2y^(2)+2x+6y+0`

Explanation

The equation of the common chord AB of the two circles is `2x+1=0` ( using `S_(1)-S_(2)=0)`.
The equation of family of circles through A and B is
`(x^(2)+y^(2)+2x+3y+1)+lambda(2x+1)=0`
[Usin `S_(1)+lambda(S_(2)=S_(1))=0]`
or `x^(2)+y^(2)+2x(lambda+1)+2y + lambda+1=0`
Since AB is a diameter of this circle, the center lies on `2x+1=0`. So,
`-2lambda-2+1=0`
or `lambda= -(1)/(2)`
Thus, the required circle is
`x^(2)+y^(2)+x+3y+(1)/(2)=0` ,brgt or `2x^(2)+2y^(2)+2x+6y+1=0`