LEARN · EXPLAIN · REVISE

Find the circle through O=(0,0), A=(2t,0), B=(0,2).

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

TOPIC 01

Equations of a circle

Construct circle equations from centers, radii, diameters and point constraints.

Construct circle equations from centers, radii, diameters and point constraints.
Justify the method and check the conditions in a new situation.

Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Transfer#Your turn

t=13

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Translate the geometric condition into a coordinate equation.

Hint 2

Use this intermediate relation.

F=0;\quad D=-2t;\quad E=-2

Worked solution

Translate the geometric condition into a coordinate equation.
Apply the stated relation and retain its conditions.
$676+26D=0\Rightarrow D=-26$
Apply the stated relation and retain its conditions.
$4+2E=0\Rightarrow E=-2$
Three noncollinear points determine one circle.

The requested relation or conclusion is shown below.

x^2+y^2-26x-2y=0

Checks and common pitfalls: Three noncollinear points determine one circle.

Reasoning checklist · self / teacher assessment

Think first. Reveal a hint when the class is ready.

Teacher preparation and assessment

Construct circle equations from centers, radii, diameters and point constraints.
Which condition is essential in equations of a circle?
Predict the shape when the constant of a circle equation changes through the zero-radius threshold.

Standard form: A circle is the locus at fixed positive distance from its center.
$(x-h)^2+(y-k)^2=r^2$
Existence: In x²+y²+Dx+Ey+F=0, a nondegenerate circle requires (D²+E²)/4−F>0.

A positive squared radius is necessary; zero gives a single point, and negative gives no real locus.