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Step 1: Define the line equation. The line $l$ passes through the point $(-1, 0)$ and has a rational slope $m$. Its equation is given by: $$y - 0 = m(x - (-1))$$ $$y = m(x + 1)$$ Step 2: Find the intersection points with the circle. Substitute the line equation into the equation of the circle $x^2 + y^2 = 1$: $$x^2 + (m(x + 1))^2 = 1$$ $$x^2 + m^2(x^2 + 2x + 1) = 1$$ $$x^2 + m^2x^2 + 2m^2x + m^2 = 1$$ $$(1 + m^2)x^2 + 2m^2x + (m^2 - 1) = 0$$ Step 3: Verify one known intersection point. The point $(-1, 0)$ is an intersection point, so $x = -1$ must be a root of the quadratic equation. Substitute $x = -1$ into the equation: $$(1 + m^2)(-1)^2 + 2m^2(-1) + (m^2 - 1) = (1 + m^2) - 2m^2 + (m^2 - 1)$$ $$= 1 + m^2 - 2m^2 + m^2 - 1 = 0$$ This confirms that $x = -1$ is indeed a root. Step 4: Determine the x-coordinate of the second intersection point P. Let $x_1 = -1$ and $x_2$ be the roots of the quadratic equation $(1 + m^2)x^2 + 2m^2x + (m^2 - 1) = 0$. By Vieta's formulas, the sum of the roots is: $$x_1 + x_2 = -\frac{2m^2}{1 + m^2}$$ Substitute $x_1 = -1$: $$-1 + x_2 = -\frac{2m^2}{1 + m^2}$$ $$x_2 = 1 - \frac{2m^2}{1 + m^2}$$ $$x_2 = \frac{1 + m^2 - 2m^2}{1 + m^2}$$ $$x_2 = \frac{1 - m^2}{1 + m^2}$$ Step 5: Confirm the rationality of $x_2$. Since $m$ is a rational number, $m^2$ is rational. Therefore, $1 - m^2$ is rational and $1 + m^2$ is rational. As $1 + m^2 \neq 0$ for any real $m$, the quotient $x_2 = \frac{1 - m^2}{1 + m^2}$ is rational. Step 6: Determine the y-coordinate of the second intersection point P. Substitute $x_2$ into the line equation $y = m(x + 1)$: $$y_2 = m\left(\frac{1 - m^2}{1 + m^2} + 1\right)$$ $$y_2 = m\left(\frac{1 - m^2 + (1 + m^2)}{1 + m^2}\right)$$ $$y_2 = m\left(\frac{2}{1 + m^2}\right)$$ $$y_2 = \frac{2m}{1 + m^2}$$ Step 7: Confirm the rationality of $y_2$. Since $m$ is a rational number, $2m$ is rational, and $1 + m^2$ is rational and non-zero. Therefore, the quotient $y_2 = \frac{2m}{1 + m^2}$ is rational. Step 8: Conclusion. The coordinates of point P are $\left(\frac{1 - m^2}{1 + m^2}, \frac{2m}{1 + m^2}\right)$. For any rational number $m$, both coordinates of P are rational. Since there are infinitely many rational numbers, there are infinitely many choices for $m$ for which both coordinates of P are rational.