The equation of the circle whose diameter is the common chord of the circles $x^2+y^2+2 x+3 y+1=0$ and $x^2+y^2+4 x+3 y+2=0$ is

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Aicharya · Accepted Answer

The equation of the common chord of the circles $x^2+y^2+2 x+3 y+1=0$ and $x^2+y^2+4 x+3 y+2=0$ is given by $ 2 x+1=0 \quad \text { [using: } S_1-S_2=0 \text { ] } $ The equation of a circle passing through the intersection of the given circles is $ \Rightarrow \begin{gathered} \left(x^2+y^2+2 x+3 y+1\right) \\ +\lambda\left(x^2+y^2+4 x+3 y+2\right)=0 \\ \quad x^2(1+\lambda)+y^2(1+\lambda)+(1+2 \lambda) \\ 2 x+3 y(1+\lambda)+1+2 \lambda=0 \end{gathered} $

Since, $2 x+1=0$ is a diameter of this circle. Therefore, its centre $\left(-\frac{2 \lambda+1}{\lambda+1},-\frac{3}{2}\right)$ lies on it $ \begin{array}{ll} \Rightarrow & -2\left(\frac{2 \lambda+1}{\lambda+1}\right)+1=0 \\ \Rightarrow & -4 \lambda-2+\lambda+1=0 \Rightarrow-3 \lambda-1=0 \\ & \lambda=-\frac{1}{3} \end{array} $ On putting $\lambda=-\frac{1}{3}$ in Eq. (i), we get $ \begin{array}{ll} \Rightarrow & x^2+y^2+\left(\frac{1-\frac{2}{3}}{1-\frac{1}{3}}\right) 2 x+3 y+\frac{1-\frac{2}{3}}{-\frac{1}{3}+1}=0 \\ \Rightarrow & x^2+y^2+\left(\frac{\frac{1}{3}}{\frac{3}{3}}\right) 2 x+3 y+\frac{\frac{1}{2}}{\frac{2}{3}}=0 \\ \Rightarrow & x^2+y^2+x+3 y+\frac{1}{2}=0 \\ \Rightarrow & 2 x^2+2 y^2+2 x+6 y+1=0 \end{array} $

The equation of the circle whose diameter is the common chord of the circles $x^2+y^2+2 x+3 y+1=0$ and…

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