Consider a circle (x-alpha)^2+(y-beta)^2=50,w | vedclass

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(D) The equation of the circle is $(x-\alpha)^2+(y-\beta)^2=50$,so the center is $C(\alpha, \beta)$ and the radius $r = \sqrt{50} = 5 \sqrt{2}$.Since

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$100$

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(D) The equation of the circle is $(x-\alpha)^2+(y-\beta)^2=50$,so the center is $C(\alpha, \beta)$ and the radius $r = \sqrt{50} = 5 \sqrt{2}$.Since the circle touches the line $x+y=0$ at point $P$,the perpendicular distance from the center $C(\alpha, \beta)$ to the line $x+y=0$ must be equal to the radius $r$.Distance $d = \frac{|\alpha+\beta|}{\sqrt{1^2+1^2}} = \frac{|\alpha+\beta|}{\sqrt{2}}$.Setting $d = r$,we get $\frac{|\alpha+\beta|}{\sqrt{2}} = 5 \sqrt{2}$.$|\alpha+\beta| = 5 \sqrt{2} 	imes \sqrt{2} = 10$.Since $\alpha, \beta > 0$,we have $\alpha+\beta = 10$.Therefore,$(\alpha+\beta)^2 = 10^2 = 100$.

Consider a circle $(x-\alpha)^2+(y-\beta)^2=50$,where $\alpha, \beta > 0$. If the circle touches the line $y+x=0$ at the point $P$,whose distance from the origin is $4 \sqrt{2}$,then $(\alpha+\beta)^2$ is equal to................

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