If frac{2}{sqrt{5}} is the length of the comm | vedclass

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(A) The equation of the common chord is given by $S_1 - S_2 = 0$.$(x^2+y^2+2x+2y+1) - (x^2+y^2+\alpha x+3y+2) = 0$$(2-\alpha)x - y - 1 = 0$.For the fi

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$4$

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(A) The equation of the common chord is given by $S_1 - S_2 = 0$.$(x^2+y^2+2x+2y+1) - (x^2+y^2+\alpha x+3y+2) = 0$$(2-\alpha)x - y - 1 = 0$.For the first circle $x^2+y^2+2x+2y+1=0$,the center is $(-1, -1)$ and the radius $r = \sqrt{(-1)^2 + (-1)^2 - 1} = 1$.The perpendicular distance $p$ from the center $(-1, -1)$ to the common chord $(2-\alpha)x - y - 1 = 0$ is:$p = \frac{|(2-\alpha)(-1) - (-1) - 1|}{\sqrt{(2-\alpha)^2 + (-1)^2}} = \frac{|\alpha-2+1-1|}{\sqrt{(2-\alpha)^2+1}} = \frac{|\alpha-2|}{\sqrt{\alpha^2-4\alpha+5}}$.The length of the common chord is $2\sqrt{r^2-p^2} = \frac{2}{\sqrt{5}}$.Thus,$\sqrt{r^2-p^2} = \frac{1}{\sqrt{5}} \Rightarrow r^2-p^2 = \frac{1}{5}$.Since $r=1$,we have $1 - p^2 = \frac{1}{5} \Rightarrow p^2 = \frac{4}{5}$.$\frac{(\alpha-2)^2}{\alpha^2-4\alpha+5} = \frac{4}{5} \Rightarrow 5(\alpha^2-4\alpha+4) = 4(\alpha^2-4\alpha+5)$.$5\alpha^2-20\alpha+20 = 4\alpha^2-16\alpha+20$.$\alpha^2 - 4\alpha = 0 \Rightarrow \alpha(\alpha-4) = 0$.Since $\alpha 
eq 0$,we get $\alpha = 4$.

If $\frac{2}{\sqrt{5}}$ is the length of the common chord of the circles $x^2+y^2+2x+2y+1=0$ and $x^2+y^2+\alpha x+3y+2=0, \alpha \neq 0$,then $\alpha=$

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