A) There cannot be any rational point on S
B) There can be infinitely many rational points on S
C) There can be at most two rational points on S
D) There are exactly two rational points on S
Correct Answer: C
Solution :
| [c] The equation of the circle S is |
| \[{{x}^{2}}+{{(y-\sqrt{2})}^{2}}={{r}^{2}}\] ? (1) |
| Let the coordinates of any point on this circle be \[(h,k),\] then |
| \[{{h}^{2}}+{{(k-\sqrt{2})}^{2}}={{r}^{2}}\] |
| \[\Rightarrow k=\sqrt{2}\pm \sqrt{{{r}^{2}}-{{h}^{2}}}\] ? (2) |
| Since the above value of k contains a constant irrational number \[\sqrt{2},\] therefore, the only possible rational value of k is 0. Hence, |
| \[\sqrt{2}\pm \sqrt{{{r}^{2}}-{{h}^{2}}}=0\Rightarrow {{r}^{2}}-{{h}^{2}}=2\] |
| \[\Rightarrow h=\pm \sqrt{{{r}^{2}}-2}\] |
| Thus, we have following cases |
| (i) if \[{{r}^{2}}-2\] is a perfect square, there will be two rational points, viz.., \[(\sqrt{{{r}^{2}}-2},0)\] and \[(-\sqrt{{{r}^{2}}-2},0)\] on S. |
| (ii) If \[{{r}^{2}}-2\] is not a perfect square, there will be no rational point on S. |
| \[\therefore \] there can be most two circle through (1, 0), |
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