x-2y-6=0 is a normal to the circle x^2+y^2+2g | vedclass

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(C) The center of the circle is $(-g, -f)$.Since the line $x-2y-6=0$ is a normal,it passes through the center:$-g - 2(-f) - 6 = 0$ $\Rightarrow -g + 2

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$4$

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(C) The center of the circle is $(-g, -f)$.Since the line $x-2y-6=0$ is a normal,it passes through the center:$-g - 2(-f) - 6 = 0$ $\Rightarrow -g + 2f = 6$ $\Rightarrow g = 2f - 6$.Given that the line $y=2$ touches the circle,the distance from the center $(-g, -f)$ to the line $y=2$ is equal to the radius $r$.$r = |-f - 2| = \sqrt{g^2 + f^2 - (-8)} = \sqrt{g^2 + f^2 + 8}$.Squaring both sides: $(f+2)^2 = g^2 + f^2 + 8$.$f^2 + 4f + 4 = g^2 + f^2 + 8 \Rightarrow 4f - 4 = g^2$.Substitute $g = 2f - 6$ into the equation:$4f - 4 = (2f - 6)^2 = 4f^2 - 24f + 36$.$4f^2 - 28f + 40 = 0 \Rightarrow f^2 - 7f + 10 = 0$.$(f-2)(f-5) = 0 \Rightarrow f = 2$ or $f = 5$.If $f = 2$,then $g = 2(2) - 6 = -2$. The radius $r = |-2 - 2| = 4$.If $f = 5$,then $g = 2(5) - 6 = 4$. The radius $r = |-5 - 2| = 7$.

$x-2y-6=0$ is a normal to the circle $x^2+y^2+2gx+2fy-8=0$. If the line $y=2$ touches this circle,then the radius of the circle can be

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