# Centre of a circle be $(-\frac{7}{10},2\sqrt2)$. Number of points of $(x,y)$ on the circle such that both $x$ and $y$ are rational

Let the coordiantes of the centre of a circle be $$(-\frac{7}{10},2\sqrt2)$$. Then the number of points of $$(x,y)$$ on the circle such that both $$x$$ and $$y$$ are rational

a.) cannot be $$3$$ or more

b.) at least $$1$$, but at most $$2$$

c.) at least $$2$$, but infinitely many

d.) infinitely many

It is an MCQ type question, only one option is correct.

My appraoch: Let us suppose we have two rational points A$$(x_1,y_1)$$ and $$B(x_2,y_2)$$ on the circle, then the perpendicular bisector of AB has equation :

$$y - \frac{y_1 + y_2}{2} = -\frac{x_2 - x_1}{y_2 - y_1}\left(x - \frac{x_1 + x_2}{2}\right)$$

Now, this line will pass through the centre $$(-\frac{7}{10},2\sqrt2)$$. Putting the value in equation we get

$$2\sqrt2= \frac{y_1 + y_2}{2} +\frac{x_2 - x_1}{y_2 - y_1}\left( \frac{7}{10}+ \frac{x_1 + x_2}{2}\right)$$

Now here we can see both points cannot be rational as left part of equation is irrational and right will become rational.

So, one point should be rational and second should be irrational. Hence there are infinitely many irrational points. Is my reasoning correct?

• You should think about the case y1=y2, too. Then you have solved the „at most“ part. However, you should also think about the existence question. Aug 26, 2020 at 15:18

You started by assuming that you had two rational points and reached a contradiction, which would show there is at most one rational point on the circle. There is a hidden assumption that $$y_1 \neq y_2$$, which allows a second point. There clearly could be no rational points, say by having the radius $$\pi$$. That says a is correct.
• There can actually be two points, if you allow $y_1=y_2$. Everything else you state is correct. Aug 26, 2020 at 15:27