Q90 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics
The equation of a circle which passes through the points of intersection of the circles $2 x^2+2 y^2-2 x+6 y-3=0$, $x^2+y^2+4 x+2 y+1=0$ and whose centre lies on the common chord of these circles is
- A$2 x^2+2 y^2-3 x+4 y-2=0$
- B$x^2+y^2+2 x+5 y-2=0$
- C$3 x^2+3 y^2-2 x+4 y-3=0$
- Correct answerD$4 x^2+4 y^2+6 x+10 y-1=0$
Solution
$\begin{aligned} & S_1 \equiv 2 x^2+2 y^2-2 x+6 y-3=0 \\ & \Rightarrow x^2+y^2-x+3 y-\frac{3}{2}=0 \\ & S_2 \equiv x^2+y^2+4 x+2 y+1=0\end{aligned}$
Eq of common chord is $S_1-S_2=0$
i.e. $-5 x+y-\frac{5}{2}=0$..(i)
Eq. of circle passing through intersections of $S_1$ and $S_2$ is
$\begin{aligned} & \Rightarrow(2+\lambda) x^2+(2+\lambda) y^2+(4 \lambda-2) x+(6+2 \lambda) y \\ & +\lambda-3=0\end{aligned}$..(ii)
$\Rightarrow x^2+y^2+\left(\frac{4 \lambda-2}{\lambda+2}\right) x+\left(\frac{2 \lambda+6}{\lambda+2}\right) y+\left(\frac{\lambda-3}{\lambda+2}\right)=0$
So, centre is $\left(-\frac{2 \lambda-1}{\lambda+2},-\frac{\lambda+3}{\lambda+2}\right)$ Centre lies in eq (i) $\Rightarrow 5\left(\frac{2 \lambda-1}{\lambda+2}\right)-\left(\frac{\lambda+3}{\lambda+2}\right)-\frac{5}{2}=0 \Rightarrow \lambda=2$ So, eq. (ii) becomes; $4 x^2+4 y^2+6 x+10 y-1=0$
So, centre is $\left(-\frac{2 \lambda-1}{\lambda+2},-\frac{\lambda+3}{\lambda+2}\right)$ Centre lies in eq (i) $\Rightarrow 5\left(\frac{2 \lambda-1}{\lambda+2}\right)-\left(\frac{\lambda+3}{\lambda+2}\right)-\frac{5}{2}=0 \Rightarrow \lambda=2$ So, eq. (ii) becomes; $4 x^2+4 y^2+6 x+10 y-1=0$
Hence, the correct answer is option: 'D'.