Q89 / 160 · TS EAMCET 2023 (12 May Shift 2) · Mathematics
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
- Correct answerA$2 x^2+2 y^2+2 x+6 y+1=0$
- B$x^2+y^2-2 x+3 y-1=0$
- C$x^2+y^2+2 x+3 y-4=0$
- D$2 x^2+2 y^2-x+2 y+1=0$
Solution
Common chord of $S_1$ and $S_2$ is $S_1-S_2=0$
$$
\begin{aligned}
& \left(x^2+y^2+2 x+3 y+1\right)-\left(x^2+y^2+4 x+3 y+2\right)=0 \\
& \Rightarrow \quad-2 x-1=0 \\
& \therefore \quad x=\frac{-1}{2}
\end{aligned}
$$
Also circle passing through $S_1$ and $S_2$
$$
\begin{aligned}
& S_1+\lambda S_2=0 \\
\Rightarrow & \left(x^2+y^2+2 x+3 y+1\right)+\lambda\left(x^2+y^2+4 x+3 y+2\right)=0 \\
\Rightarrow & (1+\lambda) x^2+(1+\lambda) y^2+(2+4 \lambda) x \\
& +(3+3 \lambda) y+(1+2 \lambda)=0 \\
\Rightarrow & x^2+y^2+\frac{(2+4 \lambda)}{\lambda+1} x+\frac{(3 \lambda+3)}{\lambda+1} y+\frac{(2 \lambda+1)}{\lambda+1}=0
\end{aligned}
$$
$\because x$ coordinate of centre is $\frac{-1}{2}$
$$
\begin{aligned}
& \Rightarrow \frac{-(1+2 \lambda)}{\lambda+1}=\frac{-1}{2} \Rightarrow 2+4 \lambda=\lambda+1 \\
& \Rightarrow 3 \lambda=-1 \Rightarrow \lambda=\frac{-1}{3}
\end{aligned}
$$
$\therefore \quad$ Equation, of circle is
$$
\begin{aligned}
& \frac{2}{3} x^2+\frac{2}{3} y^2+\frac{2}{3} x+2 y+\frac{1}{3}=0 \\
\Rightarrow & 2 x^2+2 y^2+2 x+6 y+1=0 .
\end{aligned}
$$
Hence, the correct answer is option: 'A'.