If a line along a chord of the circle 4x^{2}+ | vedclass

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(D) The equation of the circle is $4x^{2}+4y^{2}+120x+675=0$,which simplifies to $x^{2}+y^{2}+30x+\frac{675}{4}=0$.Completing the square: $(x+15)^{2}+

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$3\sqrt{5}$

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(D) The equation of the circle is $4x^{2}+4y^{2}+120x+675=0$,which simplifies to $x^{2}+y^{2}+30x+\frac{675}{4}=0$.Completing the square: $(x+15)^{2}+y^{2} = 225 - \frac{675}{4} = \frac{900-675}{4} = \frac{225}{4}$.So,the center is $(-15, 0)$ and the radius $R = \sqrt{\frac{225}{4}} = \frac{15}{2}$.The equation of a line passing through $(-30, 0)$ is $y = m(x+30)$,or $mx - y + 30m = 0$.This line is tangent to the parabola $y^{2}=30x$. The condition for tangency $y=mx+c$ to $y^{2}=4ax$ is $c = \frac{a}{m}$.Here $4a=30 \Rightarrow a = \frac{15}{2}$. The line is $y = mx + 30m$,so $c = 30m$.Thus,$30m = \frac{15/2}{m}$ $\Rightarrow 60m^{2} = 15$ $\Rightarrow m^{2} = \frac{1}{4}$ $\Rightarrow m = \pm \frac{1}{2}$.Taking $m = \frac{1}{2}$,the line is $y = \frac{1}{2}(x+30) \Rightarrow x - 2y + 30 = 0$.The perpendicular distance $P$ from the center $(-15, 0)$ to the line $x - 2y + 30 = 0$ is:$P = \frac{|-15 - 2(0) + 30|}{\sqrt{1^{2} + (-2)^{2}}} = \frac{15}{\sqrt{5}} = 3\sqrt{5}$.The length of the chord is $2\sqrt{R^{2}-P^{2}} = 2\sqrt{(\frac{15}{2})^{2} - (3\sqrt{5})^{2}} = 2\sqrt{\frac{225}{4} - 45} = 2\sqrt{\frac{225-180}{4}} = 2\sqrt{\frac{45}{4}} = \sqrt{45} = 3\sqrt{5}$.

If a line along a chord of the circle $4x^{2}+4y^{2}+120x+675=0$ passes through the point $(-30, 0)$ and is tangent to the parabola $y^{2}=30x$,then the length of this chord is:

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