Let p,q be chosen one by one from the set `{1, sqrt(2),sqrt(3), 2, e, pi}` with replacement. Now a circle is drawn taking (p,q) as its centre. Then the probability that at the most two rational points exist on the circle is (rational points are those points whose both the coordinates are rational)

Options

A. `2//3`
B. `7//8`
C. `8//9`
D. none of these

Explanation

To solve the problem, we need to determine the probability that at most two rational points exist on a circle centered at (p, q), where p and q are chosen from the set {1, √2, √3, 2, e, π}.

### Step-by-Step Solution:

1. **Identify Rational and Irrational Points:**
The set contains the following elements:
- Rational numbers: 1, 2
- Irrational numbers: √2, √3, e, π

Therefore, there are 2 rational numbers and 4 irrational numbers in the set.

2. **Total Choices for (p, q):**
Since p and q are chosen with replacement from a set of 6 elements, the total number of pairs (p, q) is:
\[ 6 \times 6 = 36 \]

3. **Count Cases for Rational Points:**
The equation of a circle centered at (p, q) is:
\[ (x - p)^2 + (y - q)^2 = r^2 \]
For rational points (x, y), both x and y must be rational.

The possible cases for the number of rational points on the circle are:
- Case 0: No rational points
- Case 1: One rational point
- Case 2: Two rational points

4. **Case Analysis:**
- **Case 0 (No rational points):** This occurs when both p and q are irrational. The number of ways to choose p and q from the irrational numbers is:
\[ 4 \times 4 = 16 \]

- **Case 1 (One rational point):** This occurs when either p or q is rational, and the other is irrational. The number of ways to choose:
- p rational and q irrational: \(2 \times 4 = 8\)
- p irrational and q rational: \(4 \times 2 = 8\)
Total for Case 1:
\[ 8 + 8 = 16 \]

- **Case 2 (Two rational points):** This occurs when both p and q are rational. The number of ways to choose p and q from the rational numbers is:
\[ 2 \times 2 = 4 \]

5. **Total Favorable Outcomes:**
The total number of favorable outcomes for having at most two rational points is:
\[ 16 \text{ (Case 0)} + 16 \text{ (Case 1)} + 4 \text{ (Case 2)} = 36 \]

6. **Calculate Probability:**
The probability that at most two rational points exist on the circle is given by the ratio of favorable outcomes to total outcomes:
\[ P(\text{at most 2 rational points}) = \frac{36}{36} = 1 \]

### Final Answer:
The probability that at most two rational points exist on the circle is \(1\).