Find rational points on the unit circle without guessing pairs of squares. Start at the known point(−1, 0) and draw a line of rational slope t. Its equation is y=t(x+1). Because one intersection with the circle is already known, solving for the other intersection requires only a linear factor after substitution. This turns a two-coordinate equation into a one-parameter family and connects rational geometry with integer right triangles.
Predict
Predict whether every slope gives distinct points and whether any point is missing. Substitution into x²+y²=1 gives (x+1)((1+t²)x+t²−1)=0. The second point is x=(1−t²)/(1+t²), y=2t/(1+t²). For real rational t the denominator is positive, so no finite parameter causes a division failure. The base point itself is not reached by any finite t; it corresponds to the vertical line through it, or an infinite projective parameter.
Experiment
Plot these points on a circle and identify the chord from the base point.
Slope t
x
y
Check x²+y²
0
1
0
1
1
0
1
1
1/2
3/5
4/5
1
−1
0
−1
1
Change the sign of t and predict which coordinate changes sign. Then choose a new rational slope, calculate both coordinates, and verify the circle equation exactly rather than with rounded decimals.
Understand
For any rational circle point other than(−1, 0), the slope t=y/(x+1) is rational. The line-intersection calculation recovers that point, proving the parametrization exhaustive apart from the stated exception. Writing t=n/m gives coordinates (m²−n²)/(m²+n²) and 2mn/(m²+n²). Clearing the denominator recovers the Pythagorean formulas. Coprimality and opposite parity are needed to classify primitive integer triples, but are not needed merely to produce rational circle points. Thus the geometric parametrization and the earlier number-theoretic classification answer related but different questions.
Apply
The mechanism extends to a nonsingular conic with rational coefficients once a rational base point is available. Substitute a rational-slope line through that point into the quadratic equation. One root is known and rational; Vieta’s relation makes the other root rational when the resulting quadratic has nonzero leading coefficient. Tangent lines may return the base point twice, a vanishing leading coefficient may put the other point at infinity, and degenerate conics require separate analysis. These exceptions are why a careful statement uses a rational parametrization with explicitly excluded parameter values rather than promising a perfect one-to-one affine formula everywhere.
Continue
A rational base point is a substantive hypothesis. The conic x²+y²=3 has real points but no rational point. If rational coordinates existed, clear denominators to a primitive integer triple a, b, c with a²+b²=3c². Modulo three, squares are zero or one, so a and b must both be divisible by three; substitution then forces c divisible by three, contradiction. A diagram of real intersections cannot remove this arithmetic obstruction. Use the slope method to construct points when a base point exists, and use congruences or descent to investigate existence when it does not.
Return to the number theory unit to choose a targeted retry. Review a new example independently before using these methods in a project.
Apply · explain · revise
Practice and reasoning
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On x²+y²=1 use slope t=2 through(−1, 0). Find x.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
For that same slope find y.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Derive the circle formulas using y=t(x+1).
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Show every rational circle point except(−1, 0) arises from a rational t.
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Prove x²+y²=3 has no rational point.
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Explain why a line of rational slope through a rational point on a rational conic generally has a rational second intersection.
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Reflect and continue
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