Question

Let C be a circle $x^2 + y^2 = 1$. The line $y = mx + m$ intersects C at the point P other than (-1,0), the number of rational choices for m for which both the coordinates of P are rational, is:

• A. 3
• B. 4
• C. 5
• D. infinitely many

Answer 1

Answer: (D)

infinitely many

Answer 2

The circle is given by the equation $C: x^2 + y^2 = 1$. The line is given by the equation $L: y = mx + m$, which can be rewritten as $y = m(x+1)$.

We substitute $y$ from the line equation into the circle equation to find the intersection points: $x^2 + (m(x+1))^2 = 1$ $x^2 + m^2(x^2 + 2x + 1) = 1$ $x^2 + m^2x^2 + 2m^2x + m^2 - 1 = 0$ $(1+m^2)x^2 + 2m^2x + (m^2 - 1) = 0$

This is a quadratic equation in $x$. We know that the line passes through $(-1,0)$ for any value of $m$. We can verify that $x=-1$ is a root: $(1+m^2)(-1)^2 + 2m^2(-1) + (m^2 - 1) = (1+m^2) - 2m^2 + m^2 - 1 = 0$. So, $x=-1$ is indeed one of the roots. Let the other root be $x_P$.

Using Vieta's formulas for the sum of roots: $-1 + x_P = -\frac{2m^2}{1+m^2}$ $x_P = -1 + \frac{2m^2}{1+m^2} = \frac{-(1+m^2) + 2m^2}{1+m^2} = \frac{m^2-1}{m^2+1}$

Now we find the y-coordinate, $y_P$, using the line equation $y_P = m(x_P+1)$: $y_P = m\left(\frac{m^2-1}{m^2+1} + 1\right) = m\left(\frac{m^2-1 + m^2+1}{m^2+1}\right) = m\left(\frac{2m^2}{m^2+1}\right) = \frac{2m^3}{m^2+1}$

So the intersection point P, other than $(-1,0)$, has coordinates $\left(\frac{m^2-1}{m^2+1}, \frac{2m^3}{m^2+1}\right)$.

The problem states that $m$ is a rational choice and asks for the number of such choices for which both coordinates of P are rational. Let $m \in \mathbb{Q}$. If $m$ is rational, then $m^2$ is rational, and $m^3$ is rational. The x-coordinate of P is $x_P = \frac{m^2-1}{m^2+1}$. If $m \in \mathbb{Q}$, then $m^2-1 \in \mathbb{Q}$ and $m^2+1 \in \mathbb{Q}$. Since $m^2 \ge 0$ for real $m$, $m^2+1 \ge 1$, so the denominator is never zero. Thus, $x_P$ is rational if $m$ is rational. The y-coordinate of P is $y_P = \frac{2m^3}{m^2+1}$. If $m \in \mathbb{Q}$, then $2m^3 \in \mathbb{Q}$ and $m^2+1 \in \mathbb{Q}$. Thus, $y_P$ is rational if $m$ is rational.

Therefore, for any rational value of $m$, both coordinates of P are rational.

We must also ensure that P is not $(-1,0)$. $x_P = \frac{m^2-1}{m^2+1} = -1 \implies m^2-1 = -(m^2+1) \implies m^2-1 = -m^2-1 \implies 2m^2 = 0 \implies m=0$. If $m=0$, the line is $y=0$. The intersection points with $x^2+y^2=1$ are $x^2=1$, so $x=\pm 1$. The points are $(-1,0)$ and $(1,0)$. In this case ($m=0$), the point P (other than $(-1,0)$) is $(1,0)$. The coordinates of P are $(1,0)$, which are rational. $m=0$ is a rational choice, and it is included.

Alternatively, we can use the standard parameterization of rational points on the unit circle starting from $(-1,0)$: $x = \frac{1-t^2}{1+t^2}, y = \frac{2t}{1+t^2}$ for $t \in \mathbb{Q}$. The line $y=m(x+1)$ has slope $m$ connecting $(-1,0)$ to the point P. If we set $t=m$, we get $x_P = \frac{1-m^2}{1+m^2}$ and $y_P = \frac{2m}{1+m^2}$. These are the coordinates of point P. If $m$ is rational, then $x_P$ and $y_P$ are rational. This parameterization generates all rational points on the circle except for $(-1,0)$. Since P is defined as the point other than $(-1,0)$, P must be one of these points generated by the parameterization. Thus, for any rational $m$, P will have rational coordinates. The set of rational numbers is infinite. Therefore, there are infinitely many rational choices for $m$.