If the circles (x-2)^2+(y-3)^2=25 and 25x^2+2 | vedclass

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(B) For the first circle $(x-2)^2+(y-3)^2=5^2$,the center $C_1 = (2, 3)$ and radius $r_1 = 5$.For the second circle $25x^2+25y^2-40x-70y-160=0$,dividi

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$-2$

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(B) For the first circle $(x-2)^2+(y-3)^2=5^2$,the center $C_1 = (2, 3)$ and radius $r_1 = 5$.For the second circle $25x^2+25y^2-40x-70y-160=0$,dividing by $25$ gives $x^2+y^2-\frac{8}{5}x-\frac{14}{5}y-\frac{32}{5}=0$.Completing the square: $(x-\frac{4}{5})^2+(y-\frac{7}{5})^2 = \frac{32}{5} + \frac{16}{25} + \frac{49}{25} = \frac{160+16+49}{25} = \frac{225}{25} = 9 = 3^2$.Thus,the center $C_2 = (\frac{4}{5}, \frac{7}{5})$ and radius $r_2 = 3$.Since the circles touch internally,the point of contact $(\alpha, \beta)$ divides the line segment joining the centers $C_1$ and $C_2$ externally in the ratio $r_1 : r_2$.$(\alpha, \beta) = \left(\frac{r_1 x_2 - r_2 x_1}{r_1 - r_2}, \frac{r_1 y_2 - r_2 y_1}{r_1 - r_2}\right) = \left(\frac{5(\frac{4}{5}) - 3(2)}{5-3}, \frac{5(\frac{7}{5}) - 3(3)}{5-3}\right) = \left(\frac{4-6}{2}, \frac{7-9}{2}\right) = (-1, -1)$.Therefore,$\alpha + \beta = -1 + (-1) = -2$.

If the circles $(x-2)^2+(y-3)^2=25$ and $25x^2+25y^2-40x-70y-160=0$ touch internally at $(\alpha, \beta)$,then $\alpha+\beta=$

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