Le n be the number of points having rational coordinates equidistant from the point `(0,sqrt3)`, the

Options

A. `ngt2`
B. `nle1`
C. `nle2`
D. `n=1`

Explanation

To solve the problem of finding the number of points with rational coordinates that are equidistant from the point (0, √3), we can follow these steps:

### Step 1: Understand the Problem
We need to find points (x, y) that are equidistant from the point (0, √3). This means that the distance from (x, y) to (0, √3) must be constant.

### Step 2: Use the Distance Formula
The distance \( d \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
For our points, we have:
\[ d = \sqrt{(x - 0)^2 + (y - \sqrt{3})^2} \]
This simplifies to:
\[ d = \sqrt{x^2 + (y - \sqrt{3})^2} \]

### Step 3: Set the Distance Equal to a Constant
Let \( r \) be the radius of the circle (which is a constant). We can set the equation as:
\[ \sqrt{x^2 + (y - \sqrt{3})^2} = r \]
Squaring both sides gives:
\[ x^2 + (y - \sqrt{3})^2 = r^2 \]

### Step 4: Expand the Equation
Expanding the left side:
\[ x^2 + (y^2 - 2y\sqrt{3} + 3) = r^2 \]
This can be rearranged to:
\[ x^2 + y^2 - 2y\sqrt{3} + 3 = r^2 \]
Thus, we have:
\[ x^2 + y^2 - 2y\sqrt{3} + 3 - r^2 = 0 \]

### Step 5: Isolate the Terms
Rearranging gives:
\[ x^2 + y^2 - 2y\sqrt{3} = r^2 - 3 \]

### Step 6: Identify Rational Coordinates
For \( (x, y) \) to have rational coordinates, \( y \) must be rational. However, \( -2y\sqrt{3} \) introduces an irrational component unless \( y = 0 \). Thus, we set:
\[ -2y\sqrt{3} = 0 \implies y = 0 \]

### Step 7: Substitute y = 0
Substituting \( y = 0 \) into the equation:
\[ x^2 + 0^2 - 2(0)\sqrt{3} = r^2 - 3 \]
This simplifies to:
\[ x^2 = r^2 - 3 \]

### Step 8: Solve for x
From this, we find:
\[ x = \pm \sqrt{r^2 - 3} \]
For \( x \) to be rational, \( r^2 - 3 \) must be a perfect square.

### Step 9: Conclusion
The number of rational points (x, y) that satisfy the condition is limited to:
- \( y = 0 \)
- \( x = \pm \sqrt{r^2 - 3} \)

Thus, there can be at most 2 points (for \( x \) being positive and negative) when \( r^2 - 3 \) is a perfect square.

### Final Answer
Therefore, the number of points \( n \) is:
\[ n \leq 2 \]