A circle is centred at $(\pi,e)$. What is the maximum no. of rational points it can have?
(A rational point is one with both coordinates rational).
1 rational point is definitely possible, just choose any rational point, and alter the radius to get it through. My book says that only one rational point is possible, as $\pi\neq qe\quad q\in Q$. That's their whole explanation. I don't understand how that's enough.
Edit:
It has been pointed out that the problem is equivalent to showing $q_1\pi+q_2e=q_3$ has no non trivial solutions. Is this known to be true? Can someone prove it in an elementary way?
[Math] Rational points on a circle with centre $(\pi, e)$
elementary-number-theorygeometry
Related Solutions
The answer is yes and $(h,k) = (a-r \sin \theta, b-r \cos \theta)$.
According to your description, the side of the triangle opposite point $(a,b)$ is parallel to the y-axis. So, start by finding the height of the triangle from point $(a,b)$, which is $h_1=r \sin \theta$. Then find the length $dy = r \cos \theta$. Works for $\theta \lt 180^\circ$.
EDIT
Just noticed I used $h$ in two different ways. Changed the height to $h_1$ in both the figure and the text.
Let the two known (shared) vertices be $A = (x_A , 0)$ and $B = (x_B , 0)$, and the two unknown vertices be $C = (x_C , y_C)$ and $D = (x_D , y_D)$.
Let the known variables be $x_A$ and $x_B$, and the distances $$\begin{aligned} L_{AB} &= x_B - x_A \gt 0 \\ L_{AC} &= \left\lVert\overline{A C}\right\rVert \gt 0 \\ L_{BC} &= \left\lVert\overline{B C}\right\rVert \gt 0 \\ L_{AD} &= \left\lVert\overline{A D}\right\rVert \gt 0 \\ L_{BD} &= \left\lVert\overline{B D}\right\rVert \gt 0 \\ \end{aligned}$$
Then, the system of equations that determines the location of $C$ and $D$ is $$\left\lbrace \begin{aligned} L_{AC}^2 &= (x_C - x_A)^2 + y_C^2 \\ L_{BC}^2 &= (x_C - x_B)^2 + y_C^2 \\ L_{AD}^2 &= (x_D - x_A)^2 + y_D^2 \\ L_{BD}^2 &= (x_D - x_B)^2 + y_D^2 \\ \end{aligned} \right .$$ This has four equations and four unknowns, and can be treated as two completely separate systems of equations, each with two unknowns ($x_C$ and $y_C$, and $x_D$ and $y_D$, respectively). The solution is $$\left\lbrace \begin{aligned} \displaystyle x_C &= \frac{ L_{AC}^2 - L_{BC}^2 + x_B^2 - x_A^2 }{2 ( x_B - x_A )} \\ \displaystyle y_C &= \pm \sqrt{ L_{AC}^2 - (x_C - x_A)^2 } = \pm \sqrt{ L_{BC}^2 - (x_C - x_B)^2 } \\ \displaystyle x_D &= \frac{ L_{AD}^2 - L_{BC}^2 + x_B^2 - x_A^2 }{2 ( x_B - x_A )} \\ \displaystyle y_D &= \pm \sqrt{ L_{AD}^2 - (x_D - x_A)^2 } = \pm \sqrt{ L_{BD}^2 - (x_D - x_B)^2 } \\ \end{aligned} \right.$$ If the two triangles extend to the same side, then we can choose the positive signs above (since $y_C \gt 0$ and $y_D \gt 0$). Both right sides for the two $y$ coordinates yield the same answer.
After solving $(x_C , y_C)$ and $(x_D , y_D)$ as above, their distance is obviously $$L_{CD} = \sqrt{ (x_D - x_C)^2 + (y_D - y_C)^2 }$$ Because the distance is necessarily nonnegative, you do not actually need the distance $L_{CD}$ itself; you can just compare the distance squared, $L_{CD}^2$, to the radius squared, $r_C^2$, because nonnegative values compare the same way (less, equal, greater) as their squares do. Expanding the above after squaring yields $$\begin{aligned} L_{CD}^2 &= \left( \frac{ L_{AC}^2 + L_{BD}^2 - L_{AD}^2 - L_{BC}^2 }{ 2 (x_B - x_A) } \right)^2 + \left( \sqrt{ L_{AC}^2 - (x_C - x_A)^2 } - \sqrt{ L_{AD}^2 - (x_D - x_A)^2 } \right)^2 \\ ~ &= \left( \frac{ L_{AC}^2 + L_{BD}^2 - L_{AD}^2 - L_{BC}^2 }{ 2 (x_B - x_A) } \right)^2 + \left( \sqrt{ L_{BC}^2 - (x_C - x_B)^2 } - \sqrt{ L_{BD}^2 - (x_D - x_B)^2 } \right)^2 \\ \end{aligned}$$ Both yield the same solution, to within numerical precision used.
Note that if you had just placed $A = (0, 0)$, ie. $x_A = 0$, you'd have gotten somewhat simpler solutions (and the math would have been easier, too).
Best Answer
Okay I got it, suppose it passes through (a,b). Then the equation of circle is $x^2-a^2+y^2-b^2-2\pi( x-a)-2e(y-b)=0$ If x and y are both rational then $q_1\pi+q_2e=q_3$ with not everything 0 .I still have to prove this impossible. I don't think it's equivalent to $\pi \neq qe$. Edit: As has been pointed out to me, two rational points are not possible if $\pi $ and $ e$ are linearly independent over the rationals, and this is still an open problem