Tangent Properties of Intersecting Circles and Parabola

To find the correct options, we analyze the given geometric configuration step-by-step.

Step 1: Point of Intersection $P$

The given equations are:

1. Circle $C_1: x^2 + y^2 = 3$
2. Parabola: $x^2 = 2y$

Substituting $x^2 = 2y$ into the equation of the circle: $2y + y^2 = 3 \implies y^2 + 2y - 3 = 0$ $(y + 3)(y - 1) = 0$

Since $P$ lies in the first quadrant, $y > 0$, so $y = 1$. Substituting $y = 1$ back into $x^2 = 2y$ gives: $x^2 = 2 \implies x = \sqrt{2} \quad (\text{since } x > 0)$

Thus, the point of intersection is $P = (\sqrt{2}, 1)$.

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Step 2: Tangent Line to Circle $C_1$ at $P$

The equation of the tangent line to the circle $x^2 + y^2 = 3$ at $P(\sqrt{2}, 1)$ is given by $x x_1 + y y_1 = 3$: $\sqrt{2}x + y = 3 \implies \sqrt{2}x + y - 3 = 0$

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Step 3: Centers of Circles $C_2$ and $C_3$

The centers $Q_2$ and $Q_3$ lie on the $y$-axis, so their coordinates can be represented as $Q(0, k)$. Since both circles $C_2$ and $C_3$ are tangent to the line $\sqrt{2}x + y - 3 = 0$ and have a radius $r = 2\sqrt{3}$, the perpendicular distance from $Q(0, k)$ to the tangent line must equal $2\sqrt{3}$:

$\frac{|\sqrt{2}(0) + k - 3|}{\sqrt{(\sqrt{2})^2 + 1^2}} = 2\sqrt{3}$ $\frac{|k - 3|}{\sqrt{3}} = 2\sqrt{3} \implies |k - 3| = 6$

This gives two possible values for $k$:

1. $k - 3 = 6 \implies k = 9 \implies Q_2 = (0, 9)$
2. $k - 3 = -6 \implies k = -3 \implies Q_3 = (0, -3)$

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Step 4: Verification of Option (A)

The distance $Q_2 Q_3$ between the centers is: $Q_2 Q_3 = |9 - (-3)| = 12$

Thus, Option (A) is correct.

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Step 5: Points of Contact $R_2$ and $R_3$ and Verification of Option (B)

$R_2$ is the foot of the perpendicular from $Q_2(0, 9)$ to the line $\sqrt{2}x + y - 3 = 0$: $\frac{x - 0}{\sqrt{2}} = \frac{y - 9}{1} = -\frac{\sqrt{2}(0) + 9 - 3}{(\sqrt{2})^2 + 1^2} = -\frac{6}{3} = -2$ $x = -2\sqrt{2}, \quad y = 9 - 2 = 7 \implies R_2 = (-2\sqrt{2}, 7)$

$R_3$ is the foot of the perpendicular from $Q_3(0, -3)$ to the line $\sqrt{2}x + y - 3 = 0$: $\frac{x - 0}{\sqrt{2}} = \frac{y - (-3)}{1} = -\frac{\sqrt{2}(0) - 3 - 3}{3} = 2$ $x = 2\sqrt{2}, \quad y = -3 + 2 = -1 \implies R_3 = (2\sqrt{2}, -1)$

Now, calculating the distance $R_2 R_3$: $R_2 R_3 = \sqrt{(2\sqrt{2} - (-2\sqrt{2}))^2 + (-1 - 7)^2}$ $R_2 R_3 = \sqrt{(4\sqrt{2})^2 + (-8)^2} = \sqrt{32 + 64} = \sqrt{96} = 4\sqrt{6}$

Thus, Option (B) is correct.

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Step 6: Verification of Option (C)

The perpendicular distance $h$ from the origin $O(0,0)$ to the tangent line $\sqrt{2}x + y - 3 = 0$ is: $h = \frac{|0 + 0 - 3|}{\sqrt{(\sqrt{2})^2 + 1^2}} = \frac{3}{\sqrt{3}} = \sqrt{3}$

The area of triangle $OR_2 R_3$ is: $\text{Area}(\triangle OR_2 R_3) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times R_2 R_3 \times h$ $\text{Area}(\triangle OR_2 R_3) = \frac{1}{2} \times 4\sqrt{6} \times \sqrt{3} = 2\sqrt{18} = 6\sqrt{2}$

Thus, Option (C) is correct.

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Step 7: Verification of Option (D)

For triangle $PQ_2 Q_3$:

• Base $Q_2 Q_3 = 12$ along the $y$-axis.
• The height is the perpendicular distance from $P(\sqrt{2}, 1)$ to the $y$-axis, which is $x_P = \sqrt{2}$.

$\text{Area}(\triangle PQ_2 Q_3) = \frac{1}{2} \times Q_2 Q_3 \times x_P = \frac{1}{2} \times 12 \times \sqrt{2} = 6\sqrt{2}$

Since Option (D) states the area is $4\sqrt{2}$, Option (D) is incorrect.

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Final Conclusion

The correct options are A, B, and C.