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 th

Correct Answer - C
Suppose, there exist three rational points or more on the circle `x^(2) + y^(2) + 2gx_(1) + 2fy + c = 0`. Therefore, if `(x_(1), y_(1)), (x_(2), y_(2))`, and `(x_(3), y_(3))` are those three points, then
`x_(1)^(2) + y_(1)^(2) + 2gx_(1) +2fy_(1) + c= 0 " (1)"`
`x_(2)^(2) + y_(2)^(2) + 2gx_(2) +2fy_(1) + c= 0 " (2)"`
`x_(3)^(2) + y_(3)^(2) + 2gx_(3) +2fy_(3) + c= 0 " (3)"`
Solving Eqs. (1), (2) and (3), we will get g, f, c as rational. Thus, center of the circle (-g, -f) is a rational point. Therefore, both the coordinates of the center are rational numbers. Obviously, the possible values of p are 1, 2. Similarly, the possible values of q are 1, 2. Thus, for this case, (p, q) may be chosen in `2 xx 2`, i.e., 4 ways. Now, (p, q) can be, without restriction, chosen in `6 xx 6`, i.e., 36 ways.
Hence, the probability that at the most rational point exist on the circle is `(36 - 4)//36 = 32//36 = 8//9`.