Let the set of all values of r, for which the | vedclass

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(A) The first circle is $C_1: (x+1)^2+(y+4)^2=r^2$ with center $O_1(-1, -4)$ and radius $r_1 = |r|$.The second circle is $C_2: x^2+y^2-4x-2y-4=0$. Rew

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$25$

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(A) The first circle is $C_1: (x+1)^2+(y+4)^2=r^2$ with center $O_1(-1, -4)$ and radius $r_1 = |r|$.The second circle is $C_2: x^2+y^2-4x-2y-4=0$. Rewriting in standard form: $(x-2)^2+(y-1)^2 = 4+1+4 = 9$, so center $O_2(2, 1)$ and radius $r_2 = 3$.For two circles to intersect at two distinct points, the distance between their centers $d = O_1O_2$ must satisfy $|r_1 - r_2| Calculate $d = \sqrt{(2 - (-1))^2 + (1 - (-4))^2} = \sqrt{3^2 + 5^2} = \sqrt{9+25} = \sqrt{34}$.Thus, $|r - 3| From $|r - 3| From $\sqrt{34}  \sqrt{34} - 3$. Since $\sqrt{34} \approx 5.83$, $\sqrt{34} - 3 > 0$, so $r > \sqrt{34} - 3$ or $r Since $r$ is a radius, $r > 0$. The intersection of $r \in (3 - \sqrt{34}, 3 + \sqrt{34})$ and $r > \sqrt{34} - 3$ is $r \in (\sqrt{34} - 3, \sqrt{34} + 3)$.Thus, $\alpha = \sqrt{34} - 3$ and $\beta = \sqrt{34} + 3$.$\alpha\beta = (\sqrt{34} - 3)(\sqrt{34} + 3) = 34 - 9 = 25$.

Let the set of all values of $r$, for which the circles $(x+1)^{2}+(y+4)^{2}=r^{2}$ and $x^{2}+y^{2}-4x-2y-4=0$ intersect at two distinct points, be the interval $(\alpha, \beta)$. Then $\alpha\beta$ is equal to

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