Conic Sections Question 219
Let S be a circle with centre $ (0,\sqrt{2}). $ Then
Options:
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
Show Answer
Answer:
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 not possible.
Hence, $ \sqrt{2}\pm \sqrt{r^{2}-h^{2}}=0\Rightarrow r^{2}-h^{2}=-2 $
$ \Rightarrow h=\pm \sqrt{r^{2}-d^{2}} $ Thus, we have the 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 at most two circles through (1, 0),
BETA
