The circle C1 :` x^2 + y^2 = 3`, with center at O, intersects the parabola` x^2 = 2y` at the point P in the first quadrant. Let the tangent to the circle C1 at P touches other two circles C2 and C3 at R2 and R3, respectively. Suppose C2 and C3 have equal radii `2sqrt(3)` and centers Q2 and Q3, respectively.If `Q_2 and Q_3` lies on the y-axis, then

Options

A. `Q_2Q_3=12`
B. `R_2R_3=4sqrt6`
C. area of `Delta OR_2R_3` is `6sqrt2`
D. area of `Delta PQ_2Q_3` is `4sqrt2`

Answer

To solve the problem step by step, we will follow the given instructions and derive the necessary equations and values.

### Step 1: Find the intersection point P of the circle and the parabola.
The equations given are:
1. Circle \( C_1: x^2 + y^2 = 3 \)
2. Parabola: \( x^2 = 2y \)

Substituting \( y \) from the parabola into the circle's equation:
\[ x^2 + \frac{x^4}{4} = 3 \]
Multiplying through by 4 to eliminate the fraction:
\[ 4x^2 + x^4 = 12 \]
Rearranging gives:
\[ x^4 + 4x^2 - 12 = 0 \]
Let \( z = x^2 \):
\[ z^2 + 4z - 12 = 0 \]
Using the quadratic formula:
\[ z = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1} = \frac{-4 \pm \sqrt{16 + 48}}{2} = \frac{-4 \pm \sqrt{64}}{2} = \frac{-4 \pm 8}{2} \]
This gives:
\[ z = 2 \quad \text{(since } z = -6 \text{ is not valid)} \]
Thus, \( x^2 = 2 \) implies \( x = \sqrt{2} \).

Now substituting back to find \( y \):
\[ y = \frac{x^2}{2} = \frac{2}{2} = 1 \]
So, the point \( P \) is \( (\sqrt{2}, 1) \).

### Step 2: Find the equation of the tangent to the circle at point P.
The slope of the radius at point \( P \) is given by:
\[ \text{slope} = -\frac{x}{y} = -\frac{\sqrt{2}}{1} = -\sqrt{2} \]
The equation of the tangent line at point \( P \) is:
\[ y - 1 = -\sqrt{2}(x - \sqrt{2}) \]
Simplifying, we get:
\[ y = -\sqrt{2}x + 2 + 1 = -\sqrt{2}x + 3 \]
Rearranging gives:
\[ \sqrt{2}x + y = 3 \]

### Step 3: Find the centers \( Q_2 \) and \( Q_3 \) of circles \( C_2 \) and \( C_3 \).
The centers \( Q_2 \) and \( Q_3 \) lie on the y-axis, so their coordinates are \( (0, y_2) \) and \( (0, y_3) \) respectively. The radius of both circles is \( 2\sqrt{3} \).

The distance from point \( P \) to the centers \( Q_2 \) and \( Q_3 \) must equal the radius of the circles. Thus, we have:
\[ \sqrt{(\sqrt{2} - 0)^2 + (1 - y_2)^2} = 2\sqrt{3} \]
Squaring both sides:
\[ 2 + (1 - y_2)^2 = 12 \]
This simplifies to:
\[ (1 - y_2)^2 = 10 \]
Taking the square root:
\[ 1 - y_2 = \pm \sqrt{10} \]
Thus:
\[ y_2 = 1 \pm \sqrt{10} \]
So, we have two possible values for \( y_2 \): \( 1 + \sqrt{10} \) and \( 1 - \sqrt{10} \).

### Step 4: Find the coordinates of \( Q_2 \) and \( Q_3 \).
Assigning:
- \( Q_2 = (0, 1 + \sqrt{10}) \)
- \( Q_3 = (0, 1 - \sqrt{10}) \)

### Step 5: Conclusion
The coordinates of the centers of circles \( C_2 \) and \( C_3 \) are:
- \( Q_2 = (0, 1 + \sqrt{10}) \)
- \( Q_3 = (0, 1 - \sqrt{10}) \)