1.

The equation of the circle whose diameter is the common chord of thecircles; `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

Answer» Here, S1 is `x^2+y^2+3x+2y+1=0`.
S2 is `x^2+y^2+3x+4y+2=0`
As both circles have common chord, we can have
`x^2+y^2+3x+2y+1 = x^2+y^2+3x+4y+2``=>y=-1/2->Eq(1)`
Now, equation of the required circle can be generate by:
`S1+lambdaS2 = 0` (From, family of circles equation)
`=>x^2+y^2+3x+2y+1+lambda(x^2+y^2+3x+4y+2) = 0`
`=>(1+lambda)x^2+(1+lambda)y^2+3(1+lambda)x+2(1+2lambda)y+(1+2lambda) = 0->Eq(2)`
Now, we can find centre of the circle from the above equation.
Centre will be `(-3/2(1+lambda),-(1+2lambda))`
As `y=-1/2`, we can say that
`-(1+2lambda) = -1/2`
`lambda = -1/4`
So, putting above value of ` lambda` in Eq(2), we get the required equation.


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