The equation of the circle whose diameter is | vedclass

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(C) Let $S_{1} \equiv x^{2}+y^{2}+2ax+c=0$ and $S_{2} \equiv x^{2}+y^{2}+2by+c=0$.The equation of the common chord is given by $S_{1}-S_{2}=0$,which i

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$x^{2}+y^{2}+\frac{2ab^{2}}{a^{2}+b^{2}}x+\frac{2a^{2}b}{a^{2}+b^{2}}y+c=0$

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(C) Let $S_{1} \equiv x^{2}+y^{2}+2ax+c=0$ and $S_{2} \equiv x^{2}+y^{2}+2by+c=0$.The equation of the common chord is given by $S_{1}-S_{2}=0$,which is $2ax-2by=0$ or $ax-by=0$.Thus,$y = \frac{ax}{b}$.Substituting $y = \frac{ax}{b}$ into $S_{1}=0$,we get $x^{2} + (\frac{ax}{b})^{2} + 2ax + c = 0$,which simplifies to $(a^{2}+b^{2})x^{2} + 2ab^{2}x + cb^{2} = 0$.Let the roots be $x_{1}, x_{2}$. Then $x_{1}+x_{2} = -\frac{2ab^{2}}{a^{2}+b^{2}}$ and $x_{1}x_{2} = \frac{cb^{2}}{a^{2}+b^{2}}$.Similarly,substituting $x = \frac{by}{a}$ into $S_{2}=0$,we get $(a^{2}+b^{2})y^{2} + 2ba^{2}y + ca^{2} = 0$.Let the roots be $y_{1}, y_{2}$. Then $y_{1}+y_{2} = -\frac{2ba^{2}}{a^{2}+b^{2}}$ and $y_{1}y_{2} = \frac{ca^{2}}{a^{2}+b^{2}}$.The equation of the circle with the common chord as diameter is $(x-x_{1})(x-x_{2}) + (y-y_{1})(y-y_{2}) = 0$.This expands to $x^{2} - (x_{1}+x_{2})x + x_{1}x_{2} + y^{2} - (y_{1}+y_{2})y + y_{1}y_{2} = 0$.Substituting the values: $x^{2} + y^{2} + \frac{2ab^{2}}{a^{2}+b^{2}}x + \frac{2a^{2}b}{a^{2}+b^{2}}y + \frac{c(b^{2}+a^{2})}{a^{2}+b^{2}} = 0$.Thus,the equation is $x^{2}+y^{2}+\frac{2ab^{2}}{a^{2}+b^{2}}x+\frac{2a^{2}b}{a^{2}+b^{2}}y+c=0$.

The equation of the circle whose diameter is the common chord of the circles $x^{2}+y^{2}+2ax+c=0$ and $x^{2}+y^{2}+2by+c=0$ is

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