If the straight line y = mx is outside the ci | vedclass

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(D) The equation of the circle is $x^2 + y^2 - 20y + 90 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 0$,$f = -10$,and $c = 90$.The

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$|m| < 3$

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(D) The equation of the circle is $x^2 + y^2 - 20y + 90 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 0$,$f = -10$,and $c = 90$.The center of the circle is $(-g, -f) = (0, 10)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{0^2 + (-10)^2 - 90} = \sqrt{100 - 90} = \sqrt{10}$.For the line $mx - y = 0$ to be outside the circle,the perpendicular distance from the center $(0, 10)$ to the line must be greater than the radius $r$.Distance $d = \frac{|m(0) - 1(10)|}{\sqrt{m^2 + (-1)^2}} = \frac{10}{\sqrt{m^2 + 1}}$.Setting $d > r$,we have $\frac{10}{\sqrt{m^2 + 1}} > \sqrt{10}$.Squaring both sides,$\frac{100}{m^2 + 1} > 10$.$10 > m^2 + 1$,which implies $m^2 Thus,$|m| < 3$.

If the straight line $y = mx$ is outside the circle $x^2 + y^2 - 20y + 90 = 0$,then

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