The equation of the circle whose diameter is the common chord of the circles $x^2+y^2+2 x+2 y+1=0$ and…
The equation of the circle whose diameter is the common chord of the circles $x^2+y^2+2 x+2 y+1=0$ and $x^2+y^2+4 x+6 y+4=0$ is
- $10 x^2+10 y^2+14 x+8 y+1=0$
- $3 x^2+3 y^2-3 x+6 y-8=0$
- $2 x^2+2 y^2-2 x+4 y+1=0$
- $x^2+y^2-x+2 y+4=0$
Solution
Given, $S_1 \equiv x^2+y^2+2 x+2 y+1=0$ and $S_2 \equiv x^2+y^2+4 x+6 y+4=0$ Now, the equation of common chord of these two circles. $S_1-S_2=0$
$
\begin{aligned}
& \left(x^2+y^2+2 x+2 y+1\right)- \\
& \quad\left(x^2+y^2+4 x+6 y+4\right)=0 \\
& \Rightarrow \quad-2 x-4 y-3=0 \text { i.e. } 2 x+4 y+3=0
\end{aligned}
$
The equation of required circle,
$
\begin{aligned}
& S_1+\lambda\left(S_2-S_1\right)= \\
& \left(x^2+y^2+2 x+2 y+1\right)+\lambda(2 x+4 y+3)=0 \\
& x^2+y^2+2 x(1+\lambda)+2 y(2 \lambda+1)+3 \lambda+1=0
\end{aligned}
$
Diameter of the above circle is the common chord of circles $S_1$ and $S_2$.
So, $2(-(1+\lambda))+4(-(2 \lambda+1))+3=0$
$
-2-2 \lambda-8 \lambda-4+3=0
$
$
10 \lambda=-3 \Rightarrow \lambda=\frac{-3}{10}
$
Hence, the required circle
$
\begin{aligned}
& x^2+y^2+2 x\left(1-\frac{3}{10}\right)+2 y\left(-\frac{6}{10}+1\right)-\frac{9}{10}+1=0 \\
& x^2+y^2+\frac{14 x}{10}+\frac{8 y}{10}+\frac{1}{10}=0 \\
& 10 x^2+10 y^2+14 x+8 y+1=0
\end{aligned}
$
Asked in: AP EAMCET 2017 (26 Apr Shift 1)
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