Circle Ques 10
- Let $C$ be any circle with centre $(0, \sqrt{2})$. Prove that at most two rational points can be there on $C$. (A rational point is a point both of whose coordinates are rational numbers.)
(1997, 5M)
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Answer:
Correct Answer: 10.(a)
Solution:
Formula:
- Equation of any circle $C$ with centre at $(0, \sqrt{2})$ is given by
or
$$ \begin{alignedat} (x-0)^{2}+(y-\sqrt{2})^{2} & =r^{2} \\ x^{2}+y^{2}-2 \sqrt{2} y+2 & =r^{2} \end{aligned} $$
where, $r>0$.
Let $\left(x _1, y _1\right),\left(x _2, y _2\right),\left(x _3, y _3\right)$ be three distinct rational points on a circle. Since a straight line parallel to $X$-axis meets a circle in at most two points, either $y _1 = y _2$ or $y _1 = y _3$.
On putting these in Eq. (i), we get
$$ \begin{alignedat} & x _1^{2}+y _1^{2}-2 \sqrt{2} y _1=r^{2}-2 \\ & x _2^{2}+y _2^{2}-2 \sqrt{2} y _2=r^{2}-2 \\ & x _3^{2}+y _3^{2}-2 \sqrt{2} y _3=r^{2}-2 \end{aligned} $$
On subtracting Eq. (ii) from Eq. (iii), we get
$$ where, \quad \begin{alignedat}{2} p _1-\sqrt{2} q _1 & =0 \\ p _1 & =x _2^{2}+y _2^{2}-x _1^{2}-y _1^{2}, \\ q _1 & =y _2-y _1 \end{aligned} $$
On subtracting Eq. (ii) from Eq. (iv), we get
$$ \text { where } \quad \begin{alignedat}{2} p _2-\sqrt{2} q _2 & =0 \\ p _2 & =x _3^{2}+y _3^{2}-x _1^{2}-y _1^{2}, q _2=y _3-y _1 \end{aligned} $$
Now, $p _1, p _2, q _1, q _2$ are rational numbers. Also, either $q _1 \neq 0$ or $q _2 \neq 0$. If $q _1 \neq 0$, then $\sqrt{2}=p _1 / q _1$ and if $q _2 \neq 0$, then $\sqrt{2}=p _2 / q _2$. In any case $\sqrt{2}$ is not a rational number. This is a contradiction.
BETA
