If the circles x^2+y^2=9 and x^2+y^2+2alpha x | vedclass

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(D) For the circle $x^2+y^2=9$,the center $C_1 = (0,0)$ and radius $r_1 = 3$. For the circle $x^2+y^2+2\alpha x+2y+1=0$,the center $C_2 = (-\alpha, -1

Vedclass · Accepted Answer

$\frac{64}{27}$

Vedclass · Answer

(D) For the circle $x^2+y^2=9$,the center $C_1 = (0,0)$ and radius $r_1 = 3$. For the circle $x^2+y^2+2\alpha x+2y+1=0$,the center $C_2 = (-\alpha, -1)$ and radius $r_2 = \sqrt{(-\alpha)^2 + (-1)^2 - 1} = \sqrt{\alpha^2} = |\alpha|$. Since the circles touch each other internally,the distance between their centers is equal to the absolute difference of their radii: $C_1C_2 = |r_1 - r_2|$. $C_1C_2 = \sqrt{(-\alpha - 0)^2 + (-1 - 0)^2} = \sqrt{\alpha^2 + 1}$. Thus,$\sqrt{\alpha^2 + 1} = |3 - |\alpha||$. Squaring both sides,$\alpha^2 + 1 = (3 - |\alpha|)^2 = 9 + \alpha^2 - 6|\alpha|$. $1 = 9 - 6|\alpha|$ $\Rightarrow 6|\alpha| = 8$ $\Rightarrow |\alpha| = \frac{4}{3}$. Therefore,$\alpha^3 = (\pm \frac{4}{3})^3 = \pm \frac{64}{27}$. Given the options,we consider $\alpha = \frac{4}{3}$,so $\alpha^3 = \frac{64}{27}$.

If the circles $x^2+y^2=9$ and $x^2+y^2+2\alpha x+2y+1=0$ touch each other internally,then the value of $\alpha^3$ is

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