Consider L_1: 2x + 3y + p - 3 = 0; L_2: 2x + | vedclass

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(C) The equation of the circle $C$ is $x^2 + 6x + 9 + y^2 - 10y + 25 = -30 + 9 + 25$,which simplifies to $(x + 3)^2 + (y - 5)^2 = 4$.The center of the

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$STATEMENT-1$ is True,$STATEMENT-2$ is False.

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(C) The equation of the circle $C$ is $x^2 + 6x + 9 + y^2 - 10y + 25 = -30 + 9 + 25$,which simplifies to $(x + 3)^2 + (y - 5)^2 = 4$.The center of the circle is $(-3, 5)$ and the radius $r = 2$.For $L_1: 2x + 3y + p - 3 = 0$ to be a chord,the distance from the center $(-3, 5)$ to $L_1$ must be less than the radius $r=2$.Distance $d_1 = \frac{|2(-3) + 3(5) + p - 3|}{\sqrt{2^2 + 3^2}} = \frac{|p + 6|}{\sqrt{13}} For $L_2: 2x + 3y + p + 3 = 0$ to be a diameter,it must pass through the center $(-3, 5)$,so $2(-3) + 3(5) + p + 3 = 0 \Rightarrow p = -12$.If $p = -12$,$L_1$ becomes $2x + 3y - 15 = 0$. The distance $d_1 = \frac{|-12 + 6|}{\sqrt{13}} = \frac{6}{\sqrt{13}} \approx 1.66 For $STATEMENT-2$: If $L_1$ is a diameter,$2(-3) + 3(5) + p - 3 = 0 \Rightarrow p = -6$. Then $L_2$ is $2x + 3y - 3 = 0$. The distance from the center to $L_2$ is $d_2 = \frac{|2(-3) + 3(5) - 3|}{\sqrt{13}} = \frac{6}{\sqrt{13}} \approx 1.66 < 2$. Since $d_2 < r$,$L_2$ is a chord. Thus,$STATEMENT-2$ is False.

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