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If $\frac{2}{\sqrt{5}}$ is the length of the common chord of the circles $x^2+y^2+2 x+2 y+1=0$ and $x^2+y^2+\alpha x+3 y+2=0, \alpha \neq 0$, then $\alpha=$
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Verified Answer
The correct answer is:
4
Equation of common chord $=S_1-S_2=0$
$$
\begin{array}{r}
{\left[x^2+y^2+2 x+2 y+1\right]-\left[x^2+y^2+\alpha x+3 y+2\right]=0} \\
(2-\alpha) x-y-1=0
\end{array}
$$
Length of common chord $=2 \sqrt{r^2-p^2}$
$$
\begin{aligned}
& p=\frac{(2-\alpha)(-1)-(-1)-1}{\sqrt{(2-\alpha)^2+1}} \Rightarrow \frac{\alpha-2+1-1}{\sqrt{4+\alpha^2-4 \alpha+1}} \\
\Rightarrow \quad p^2=\frac{\alpha^2-4 \alpha+4}{\alpha^2-4 \alpha+5} \quad \Rightarrow & 2 \sqrt{r^2-p^2}=\frac{2}{\sqrt{5}} \\
\Rightarrow \quad \sqrt{r^2-p^2}=\frac{1}{\sqrt{5}} \quad \Rightarrow & r^2-p^2=\frac{1}{5}=1 \\
\Rightarrow \quad & \quad[\because r=1] \\
\Rightarrow \quad & 5-5\left(\frac{\alpha^2-4 \alpha+4}{\alpha^2-4 \alpha+5}\right)=1
\end{aligned}
$$
$$
\begin{array}{lrlrl}
& \Rightarrow & 5 \alpha^2-20 \alpha+25-5 \alpha^2+20 \alpha-20=\alpha^2-4 \alpha+5 \\
& \Rightarrow & & \alpha^2-4 \alpha+5 & =5 \\
& \Rightarrow & \alpha(\alpha-4) & =0 \\
& \therefore & \alpha & = & 4
\end{array}
$$
$$
\begin{array}{r}
{\left[x^2+y^2+2 x+2 y+1\right]-\left[x^2+y^2+\alpha x+3 y+2\right]=0} \\
(2-\alpha) x-y-1=0
\end{array}
$$
Length of common chord $=2 \sqrt{r^2-p^2}$
$$
\begin{aligned}
& p=\frac{(2-\alpha)(-1)-(-1)-1}{\sqrt{(2-\alpha)^2+1}} \Rightarrow \frac{\alpha-2+1-1}{\sqrt{4+\alpha^2-4 \alpha+1}} \\
\Rightarrow \quad p^2=\frac{\alpha^2-4 \alpha+4}{\alpha^2-4 \alpha+5} \quad \Rightarrow & 2 \sqrt{r^2-p^2}=\frac{2}{\sqrt{5}} \\
\Rightarrow \quad \sqrt{r^2-p^2}=\frac{1}{\sqrt{5}} \quad \Rightarrow & r^2-p^2=\frac{1}{5}=1 \\
\Rightarrow \quad & \quad[\because r=1] \\
\Rightarrow \quad & 5-5\left(\frac{\alpha^2-4 \alpha+4}{\alpha^2-4 \alpha+5}\right)=1
\end{aligned}
$$
$$
\begin{array}{lrlrl}
& \Rightarrow & 5 \alpha^2-20 \alpha+25-5 \alpha^2+20 \alpha-20=\alpha^2-4 \alpha+5 \\
& \Rightarrow & & \alpha^2-4 \alpha+5 & =5 \\
& \Rightarrow & \alpha(\alpha-4) & =0 \\
& \therefore & \alpha & = & 4
\end{array}
$$
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