圆的方程yuán de fāngchéng

Core Concept

A circle is the set of all points at a fixed distance $r$ (the radius, 半径) from a fixed point $C(a, b)$ (the centre, 圆心). Because that definition is a statement about distance, the equation of a circle is nothing more than the distance formula with both sides squared.

Standard Form (标准方程)

$(x - a)^2 + (y - b)^2 = r^2$

• Centre: $(a, b)$
• Radius: $r$

Read the signs carefully. The equation $(x - 3)^2 + (y + 2)^2 = 25$ has centre $(3, -2)$, not $(3, 2)$ — the standard form subtracts the centre's coordinates, so a $+2$ inside means the centre coordinate is $-2$.

When the centre is the origin, the form collapses to $x^2 + y^2 = r^2$.

General Form (一般方程)

Expanding the standard form and relabelling gives:

$x^2 + y^2 + Dx + Ey + F = 0$

From it you recover:

$\text{centre} = \left(-\frac{D}{2}, -\frac{E}{2}\right), \qquad r = \frac{1}{2}\sqrt{D^2 + E^2 - 4F}$

When Is It Actually a Circle?

Not every equation of that shape is a circle. Let $\Delta = D^2 + E^2 - 4F$:

Condition	Graph
$\Delta > 0$	a genuine circle
$\Delta = 0$	a single point (点圆)
$\Delta < 0$	no real graph at all

This classification is a standard CSCA question — usually phrased as "for what values of $m$ does this equation represent a circle?"

Converting General → Standard

Use completing the square (配方法) on each variable separately. For $x^2 + y^2 - 6x + 4y - 12 = 0$:

1. Group: $(x^2 - 6x) + (y^2 + 4y) = 12$
2. Complete each square: $(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4$
3. Factor: $(x-3)^2 + (y+2)^2 = 25$

Centre $(3, -2)$, radius $5$. Note that whatever you add on the left must be added on the right too.

Line and Circle: Positional Relationships

Let $d$ be the distance from the centre to the line (use the point-to-line distance formula) and $r$ the radius:

Relationship	Condition	Intersection points
Separate 相离	$d > r$	0
Tangent 相切	$d = r$	1
Intersecting 相交	$d < r$	2

Comparing $d$ with $r$ is almost always faster than substituting the line into the circle and computing a discriminant — this is the single biggest time-saver on circle questions.

Chord Length

If a line cuts a chord of length $L$ across a circle of radius $r$ at distance $d$ from the centre:

$L = 2\sqrt{r^2 - d^2}$

This is the Pythagorean theorem applied to half the chord, the radius, and the perpendicular from the centre.

CSCA Practice Problems

💡 Note: These problems follow the CSCA exam syllabus and Chinese standardized test formats.

Example 1: Basic (Difficulty ★★☆☆☆)

The circle $x^2 + y^2 - 4x + 6y - 3 = 0$ has centre and radius:

Options:

• A. centre $(2, -3)$, radius $4$
• B. centre $(-2, 3)$, radius $4$
• C. centre $(2, -3)$, radius $16$
• D. centre $(-2, 3)$, radius $16$

Detailed Solution:

Apply the general-form formulas with $D = -4$, $E = 6$, $F = -3$:

• Centre: $\left(-\frac{-4}{2}, -\frac{6}{2}\right) = (2, -3)$
• Radius: $r = \frac{1}{2}\sqrt{(-4)^2 + 6^2 - 4(-3)} = \frac{1}{2}\sqrt{16 + 36 + 12} = \frac{1}{2}\sqrt{64} = 4$

Option C is the trap for students who compute $r^2$ and forget the final square root.

Answer: A

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Example 2: Intermediate (Difficulty ★★★☆☆)

The line $3x + 4y - 5 = 0$ intersects the circle $x^2 + y^2 = 4$ in a chord. Find the chord's length.

Options:

• A. $2\sqrt{2}$
• B. $2\sqrt{3}$
• C. $3$
• D. $2$

Detailed Solution:

1. The circle has centre $(0, 0)$ and radius $r = 2$.

2. Distance from the centre to the line: $d = \frac{|3(0) + 4(0) - 5|}{\sqrt{3^2 + 4^2}} = \frac{5}{5} = 1$

3. Since $d = 1 < 2 = r$, the line does intersect the circle in two points. ✓

4. Chord length: $L = 2\sqrt{r^2 - d^2} = 2\sqrt{4 - 1} = 2\sqrt{3}$

Answer: B

Why not substitute? Substituting the line into the circle gives a quadratic whose roots you would then feed into the distance formula — perhaps two minutes of work. The $d$-versus-$r$ route above takes twenty seconds.

Common Mistakes

❌ Mistake 1: Sign errors reading the centre

$(x + 1)^2 + (y - 4)^2 = 9$ has centre $(-1, 4)$. The standard form is written with minus signs, so the sign you see inside the bracket is the opposite of the centre's coordinate.

❌ Mistake 2: Reporting $r^2$ instead of $r$

The right-hand side of the standard form is $r^2$. If it reads $25$, the radius is $5$.

❌ Mistake 3: Forgetting to balance when completing the square

Adding $9$ inside $(x^2 - 6x + 9)$ on the left requires adding $9$ on the right. Skipping this shifts the radius and quietly ruins every subsequent step.

❌ Mistake 4: Assuming any $x^2 + y^2 + Dx + Ey + F = 0$ is a circle

Always check $D^2 + E^2 - 4F > 0$ first, especially when the equation contains a parameter.

Study Tips

1. ✅ Remember the circle equation is just the distance formula squared — you can re-derive it under pressure
2. ✅ Compare $d$ with $r$ for line-circle problems instead of substituting
3. ✅ Practice completing the square until it is automatic in both variables
4. ✅ Memorise the chord formula $L = 2\sqrt{r^2 - d^2}$ — it turns a long problem into one line
5. ✅ Check the $\Delta > 0$ condition whenever a parameter appears

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💡 Exam Tip: 由圆心（或半径）求圆的方程 and 由标准方程确定圆心和半径 together make circles the second-most-tested analytic-geometry topic in CSCA, behind only the distance formula. Most circle questions are two steps: find the centre and radius, then apply one positional-relationship rule.