Standard form
A circle is the locus at fixed positive distance from its center.
LEARN · EXPLAIN · REVISE
Read the idea, work independently, then explain what changed.
高二選擇性必修 第一册(A版).pdf · 2.4 · PDF 87 / printed page 82
Revisit first: Intersections and distance formulas
TOPIC 01
Construct circle equations from centers, radii, diameters and point constraints.
A circle is the locus at fixed positive distance from its center.
In x²+y²+Dx+Ey+F=0, a nondegenerate circle requires (D²+E²)/4−F>0.
PREDICT → EXPLORE → EXPLAIN → TRANSFER
Lesson question: Predict the shape when the constant of a circle equation changes through the zero-radius threshold.
Model exploration: predict which displayed result changes with the parameters, check the values, and compare the observation with the lesson question.
Circle x²+y²=4; line y=1. 2 intersections; chord length=3.4641. Tangency occurs exactly when h=r.
Explain: Calculate two valid cases and explain the change using the defining relation.
Transfer: Compare a circle, a point and an empty real locus by completing squares.
Use one hint at a time. A correction explains what changed, not just the final answer.
Working and explanation
BUILD THE REASONING
Complete squares before reading the geometric parameters.
Apply the stated relation and retain its conditions.
The radius is the positive square root.
The requested value is 2.
Checks and common pitfalls: The radius is the positive square root.
Think first. Reveal a hint when the class is ready.
Working and explanation
BUILD THE REASONING
Complete squares before reading the geometric parameters.
Apply the stated relation and retain its conditions.
Apply the stated relation and retain its conditions.
The zero-radius boundary is a point rather than a circle.
The requested relation or conclusion is shown below.
Checks and common pitfalls: The zero-radius boundary is a point rather than a circle.
Think first. Reveal a hint when the class is ready.
Working and explanation
BUILD THE REASONING
Translate the geometric condition into a coordinate equation.
Apply the stated relation and retain its conditions.
The extremal point lies on the horizontal diameter.
The requested value is 6.
Checks and common pitfalls: The extremal point lies on the horizontal diameter.
Think first. Reveal a hint when the class is ready.
Working and explanation
BUILD THE REASONING
Complete squares before reading the geometric parameters.
Apply the stated relation and retain its conditions.
The radius is the positive square root.
The requested value is 5.
Checks and common pitfalls: The radius is the positive square root.
Think first. Reveal a hint when the class is ready.
Working and explanation
BUILD THE REASONING
Complete squares before reading the geometric parameters.
Apply the stated relation and retain its conditions.
The center has the opposite signs from the offsets inside the squares.
The requested value is 6.
Checks and common pitfalls: The center has the opposite signs from the offsets inside the squares.
Think first. Reveal a hint when the class is ready.
Working and explanation
BUILD THE REASONING
Translate the geometric condition into a coordinate equation.
Apply the stated relation and retain its conditions.
Apply the stated relation and retain its conditions.
The radius is half the diameter, so its square is one quarter.
The requested value is 53.
Checks and common pitfalls: The radius is half the diameter, so its square is one quarter.
Think first. Reveal a hint when the class is ready.
Working and explanation
BUILD THE REASONING
Translate the geometric condition into a coordinate equation.
Apply the stated relation and retain its conditions.
Use the center-to-point distance; the point itself is not the center.
The requested relation or conclusion is shown below.
Checks and common pitfalls: Use the center-to-point distance; the point itself is not the center.
Think first. Reveal a hint when the class is ready.
Working and explanation
BUILD THE REASONING
Translate the geometric condition into a coordinate equation.
Apply the stated relation and retain its conditions.
Use the center-to-point distance; the point itself is not the center.
The requested relation or conclusion is shown below.
Checks and common pitfalls: Use the center-to-point distance; the point itself is not the center.
Think first. Reveal a hint when the class is ready.
Working and explanation
BUILD THE REASONING
Complete squares before reading the geometric parameters.
Apply the stated relation and retain its conditions.
Apply the stated relation and retain its conditions.
The zero-radius boundary is a point rather than a circle.
The requested relation or conclusion is shown below.
Checks and common pitfalls: The zero-radius boundary is a point rather than a circle.
Think first. Reveal a hint when the class is ready.
Working and explanation
BUILD THE REASONING
Translate the geometric condition into a coordinate equation.
Apply the stated relation and retain its conditions.
Apply the stated relation and retain its conditions.
Three noncollinear points determine one circle.
The requested relation or conclusion is shown below.
Checks and common pitfalls: Three noncollinear points determine one circle.
Think first. Reveal a hint when the class is ready.
Working and explanation
BUILD THE REASONING
Translate the geometric condition into a coordinate equation.
Apply the stated relation and retain its conditions.
The extremal point lies on the horizontal diameter.
The requested value is 14.
Checks and common pitfalls: The extremal point lies on the horizontal diameter.
Think first. Reveal a hint when the class is ready.
Working and explanation
BUILD THE REASONING
Translate the geometric condition into a coordinate equation.
Apply the stated relation and retain its conditions.
Apply the stated relation and retain its conditions.
Three noncollinear points determine one circle.
The requested relation or conclusion is shown below.
Checks and common pitfalls: Three noncollinear points determine one circle.
Think first. Reveal a hint when the class is ready.
Working and explanation
BUILD THE REASONING
Translate the geometric condition into a coordinate equation.
Apply the stated relation and retain its conditions.
The extremal point lies on the horizontal diameter.
The requested value is 16.
Checks and common pitfalls: The extremal point lies on the horizontal diameter.
Think first. Reveal a hint when the class is ready.
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