Number of irrational coordinates of the circle $x^2+y^2=5$.

Question

It is given in the book by Malcolm Cameron, titled: Mathematics the Truth; that the circle $x^2+y^2=5$ has an infinite number of points with at least one coordinate irrational, but also an infinite number of rational coordinates too.
I cannot understand. Starting with the latter part as only $(pm 1, pm 2)$ seem to be only the $4$ points with rational coordinates; rather than an infinite number of rational coordinates.
Similarly for the irrational coordinates, as not clear how an infinite number of such (at least one) coordinates can be there. Any such points have to anyway satisfy $sqrt{a^2+b^2}=5,  a,bin mathbb{R}$.
So, how to prove this statement.
I request at least suitable hints in the direction to pursue.

Angina Seng · Answer

Consider the point $P=(2,1)$ on the circle. Take any line through $P$ with rational gradient:
$$y-1=m(x-2)$$
with $minBbb Q$. This meets the circle in two points (unless it's the tangent
at $P$), namely $P$ and $Q$ say. Then $Q$ also has rational coordinates.
As an example take $m=2$. Then the line is $y=2x-3$. Substituting in $x^2+y^2=5$
gives
$$x^2+(2x-3)^2=5x^2-12x+9=5.$$
This quadratic has roots $x=2$ and $x=2/5$ leading to $P=(2,1)$ and $Q=(2/5,-11/5)$.

Number of irrational coordinates of the circle $x^2+y^2=5$.

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