The circle C_1: x^2+y^2=3,with centre at O,in | vedclass

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(A, B, C) The circle is $x^2+y^2=3$ and the parabola is $x^2=2y$. Substituting $x^2=2y$ into the circle equation gives $2y+y^2=3$,so $y^2+2y-3=0$,whic

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$A, B, D$

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(A, B, C) The circle is $x^2+y^2=3$ and the parabola is $x^2=2y$. Substituting $x^2=2y$ into the circle equation gives $2y+y^2=3$,so $y^2+2y-3=0$,which factors as $(y+3)(y-1)=0$. Since $P$ is in the first quadrant,$y=1$,and $x^2=2(1)=2$,so $x=\sqrt{2}$. Thus,$P \equiv (\sqrt{2}, 1)$.The tangent to $x^2+y^2=3$ at $(\sqrt{2}, 1)$ is $\sqrt{2}x + y = 3$. Let this line be $L$. The slope of $L$ is $m = -\sqrt{2}$.Let $	heta$ be the angle the line $L$ makes with the positive $x$-axis,so $	an 	heta = -\sqrt{2}$. The angle $\alpha$ between the line $L$ and the $y$-axis satisfies $	an \alpha = |\cot 	heta| = \frac{1}{|	an 	heta|} = \frac{1}{\sqrt{2}}$.Let $T$ be the intersection of $L$ with the $y$-axis. Setting $x=0$ in $\sqrt{2}x+y=3$ gives $T \equiv (0, 3)$.For circles $C_2, C_3$ with radius $r=2\sqrt{3}$ tangent to $L$ at $R_2, R_3$ with centers $Q_2, Q_3$ on the $y$-axis,the distance $Q_3T = \frac{r}{\sin \alpha}$. Since $	an \alpha = \frac{1}{\sqrt{2}}$,$\sin \alpha = \frac{1}{\sqrt{3}}$. Thus $Q_3T = \frac{2\sqrt{3}}{1/\sqrt{3}} = 6$. Since $Q_2, Q_3$ are symmetric about $T$ on the $y$-axis,$Q_2Q_3 = 2 	imes 6 = 12$. (Option $A$ is correct).$R_3T = \frac{r}{	an \alpha} = \frac{2\sqrt{3}}{1/\sqrt{2}} = 2\sqrt{6}$. Thus $R_2R_3 = 2 	imes 2\sqrt{6} = 4\sqrt{6}$. (Option $B$ is correct).The perpendicular distance from $O(0,0)$ to $L$ is $d = \frac{|3|}{\sqrt{(\sqrt{2})^2+1^2}} = \frac{3}{\sqrt{3}} = \sqrt{3}$. Area of $	riangle OR_2R_3 = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (4\sqrt{6}) 	imes \sqrt{3} = 2\sqrt{18} = 6\sqrt{2}$. (Option $C$ is correct).Area of $	riangle PQ_2Q_3 = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (Q_2Q_3) 	imes |x_P| = \frac{1}{2} 	imes 12 	imes \sqrt{2} = 6\sqrt{2}$. (Option $D$ is incorrect).

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