The equation of the circle whose diameter is the common chord of the circles; `x^2+y^2+3x+2y+1=0` & `x^2+y^2+3x+4y+2=0` is: `x^2+y^2+8x+10 y+2=0` `x^2+y^2-5x+4y+7=0` `2x^2+2y^2+6x+2y+1=0` None of these

Options

A. `x^(2)+y^(2)+8x+10y+2=0`
B. `x^(2)+y^(2)-5x+4y+7=0`
C. `2x^(2)+2y^(2)+6x+2y+1=0`
D. none of these

Answer

C

Explanation

The equation of the common chord of the circles
`x^(2)+y^(2)+3x+2y+1=0 and x^(2)+y^(2)+3x+4y+2=0` is given by
2y+1=0 [Using : `S_(1)-S_(2)=0`]
The equation of a circle passing through the intersection of the given circles is
`(x^(2)+y^(2)+3x+2y+1)+lambda(x^(2)+y^(2)+3x+4y+2)=0`
`rArr x^(2)+y^(2)+x((3+3lambda)/(lambda+1))+2lambda((1+2lambda)/(lambda+1))+(1+2lambda)/(lambda+1)=0` ...(i)
Since 2 y+1=0 is a diameter of this circle. Therefore, its centre
`(-(3)/(2), -(2lambda+1)/(lambda + 1))` lies on it.
`:. - 2 ((2lambda+1)/(lambda+1))+1=0rArr - 4 lambda-2 + lambda+1=0rArr lambda=-(1)/(3)`
Putting `lambda=-1//3` in (i), the equation of the required circle is
`2x^(2)+2y^(2)+6x+2y+1=0`