Unit 8 · Circles　圆

By the end of this chapter you can:

1. Write the standard equation of a circle given its centre and radius, and read both back from the equation
2. Convert the general equation $x^{2}+y^{2}+Dx+Ey+F=0$ to standard form by completing the square, and decide whether it represents a real circle
3. Determine the position relationship between a line and a circle by comparing the centre-to-line distance $d$ with the radius $r$
4. Find the equation of a tangent line to a circle, both at a given point on the circle and from an external point

Exam weight on past CSCA papers: ~5% (2 of 48 MCQs).

8.1　Standard Equation of a Circle　圆的标准方程

A circle is the set of all points in the plane that are a fixed distance $r$ (the radius) from a fixed point $C(a, b)$ (the centre). Applying the distance formula, any point $P(x, y)$ on the circle satisfies

$$\boxed{(x - a)^{2} + (y - b)^{2} = r^{2}}$$

Special cases worth knowing:

• Centre at the origin: $x^{2} + y^{2} = r^{2}$.
• Unit circle: $x^{2} + y^{2} = 1$ (centre $O$, radius $1$).

⚠️ Sign trap: The standard form is $(x - a)^{2} + (y - b)^{2} = r^{2}$. A circle centred at $(3, -2)$ is $(x - 3)^{2} + (y + 2)^{2} = r^{2}$ — the sign on $y$ flips because $y - (-2) = y + 2$. Students who rush will write $(y - 2)^{2}$ and get the wrong centre.

🔑 Reading a standard equation: identify what fills $\square$ in $(x - \square)^{2} + (y - \square)^{2} = \square$. The centre is $(\square, \square)$ and the radius is $\sqrt{\square}$.

Worked Example 8.1.A

Write the equation of the circle with centre $C(-1, 3)$ and radius $r = 5$. Then verify that $P(3, 6)$ lies on the circle.

Solution.

Substitute directly into the standard form:

$$(x - (-1))^{2} + (y - 3)^{2} = 5^{2}$$

$$\boxed{(x + 1)^{2} + (y - 3)^{2} = 25}$$

Verification — plug $P(3, 6)$ into the left side:

$$(3 + 1)^{2} + (6 - 3)^{2} = 16 + 9 = 25\ ✓$$

8.2　General Equation of a Circle　圆的一般方程

Expanding the standard form $(x-a)^{2}+(y-b)^{2}=r^{2}$ and rearranging gives the general equation:

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

where $D = -2a$, $E = -2b$, $F = a^{2} + b^{2} - r^{2}$.

Reading back the centre and radius using the coefficients directly:

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

Validity condition: The equation represents a real circle if and only if

$$D^{2} + E^{2} - 4F > 0$$

(If $D^{2} + E^{2} - 4F = 0$ you get a single point; if $< 0$, there is no real geometric locus.)

How to complete the square (the reliable method):

1. Group the $x$-terms and $y$-terms separately, moving the constant to the right.
2. For $x^{2} + Dx$: add $\left(\tfrac{D}{2}\right)^{2}$ to both sides to get $\left(x + \tfrac{D}{2}\right)^{2}$.
3. Repeat for $y^{2} + Ey$.
4. The right side is now $r^{2}$; read off centre and radius.

⚠️ Complete the square — don't shortcut: applying the centre formula $(-D/2,\,-E/2)$ is fast but skips the validity check. Always verify $D^{2}+E^{2}-4F>0$ before claiming the equation is a circle.

🔑 Quick check after completing the square: the right-hand side you end up with must be positive. If it's zero or negative, it is not a real circle — the CSCA will use this as a distractor.

Worked Example 8.2.A

Convert $x^{2} + y^{2} - 6x + 4y - 3 = 0$ to standard form. Find the centre and radius.

Solution.

Rearrange and group:

$$(x^{2} - 6x) + (y^{2} + 4y) = 3$$

Complete the square for $x$: add $\left(\tfrac{6}{2}\right)^{2} = 9$ to both sides.

Complete the square for $y$: add $\left(\tfrac{4}{2}\right)^{2} = 4$ to both sides.

$$(x^{2} - 6x + 9) + (y^{2} + 4y + 4) = 3 + 9 + 4 = 16$$

$$(x - 3)^{2} + (y + 2)^{2} = 16$$

$$\boxed{\text{Centre } (3,\ -2),\quad r = 4}$$

Validity check: $D = -6$, $E = 4$, $F = -3$. $\ D^{2}+E^{2}-4F = 36 + 16 + 12 = 64 > 0$. ✓

8.3　Line–Circle Position Relationships　直线与圆的位置关系

Given a circle with centre $C(a, b)$ and radius $r$, and a line $\ell: Ax + By + C_{0} = 0$, let

$$d = \frac{|Aa + Bb + C_{0}|}{\sqrt{A^{2} + B^{2}}}$$

be the perpendicular distance from the centre $C$ to the line $\ell$. The three cases are:

Chord length formula: when the line is a secant ($d < r$), the perpendicular from the centre bisects the chord. By the Pythagorean theorem the half-chord has length $\sqrt{r^{2} - d^{2}}$, so the full chord length is

$$\ell_{\text{chord}} = 2\sqrt{r^{2} - d^{2}}$$

🔑 $d$ vs $r$ — the whole game: memorise this table. On the CSCA you compute $d$ once and compare:

• $d < r$: line cuts through — two points, chord can be calculated.
• $d = r$: line just grazes — one tangent point.
• $d > r$: line misses entirely — zero points.

Worked Example 8.3.A

Determine the relationship between the circle $(x - 1)^{2} + (y + 2)^{2} = 25$ and the line $3x - 4y + 15 = 0$. If they intersect, find the chord length.

Solution.

Centre $C(1, -2)$, radius $r = 5$.

$$d = \frac{|3(1) - 4(-2) + 15|}{\sqrt{3^{2} + (-4)^{2}}} = \frac{|3 + 8 + 15|}{5} = \frac{26}{5} = 5.2$$

Since $d = 5.2 > r = 5$, the line and circle are disjoint (相离).

$$\boxed{\text{No intersection. The line misses the circle.}}$$

(If the answer had given $d = 3 < 5$, the chord length would have been $2\sqrt{25 - 9} = 2\sqrt{16} = 8$.)

8.4　Tangent Lines to a Circle　圆的切线

Two standard setups appear on the CSCA.

Setup 1 — Tangent at a known point on the circle

If the circle is $(x - a)^{2} + (y - b)^{2} = r^{2}$ and you need the tangent at the point $(x_{0},\ y_{0})$ on the circle, the tangent line is

$$\boxed{(x_{0} - a)(x - a) + (y_{0} - b)(y - b) = r^{2}}$$

Why it works: the radius from $C(a,b)$ to $(x_{0},y_{0})$ is perpendicular to the tangent. The formula above is the dot-product condition that the tangent vector is perpendicular to the radius vector.

Special case (circle centred at origin): tangent at $(x_{0}, y_{0})$ on $x^{2}+y^{2}=r^{2}$ is

$$x_{0}x + y_{0}y = r^{2}$$

Setup 2 — Tangent from an external point

If $P(x_{1}, y_{1})$ lies outside the circle (verify: $|PC|^{2} > r^{2}$), there are exactly two tangent lines from $P$ to the circle.

Tangent length (the common distance from $P$ to either contact point $T$):

$$|PT| = \sqrt{|PC|^{2} - r^{2}}$$

To find the tangent line equations:

1. Write the family of lines through $P$: $y - y_{1} = k(x - x_{1})$, i.e. $kx - y + (y_{1} - kx_{1}) = 0$.
2. Set the centre-to-line distance equal to $r$ and solve for $k$.
3. Separately check whether the vertical line $x = x_{1}$ is also tangent.

⚠️ Don't forget the vertical tangent: the slope $k$ approach misses vertical lines. Always test $x = x_{1}$ as a separate candidate — compute $d$ and compare to $r$.

🔑 Two-step strategy for external-point tangents:

1. Confirm $P$ is external: $|PC|^{2} > r^{2}$.
2. Set up $y - y_{1} = k(x - x_{1})$, apply $d = r$, solve the resulting quadratic in $k$.

Worked Example 8.4.A

Find the equations of the tangent lines from $P(3, 1)$ to the circle $x^{2} + y^{2} = 5$.

Solution.

Step 1 — confirm $P$ is external.

Circle: centre $C(0, 0)$, radius $r = \sqrt{5}$. $|PC|^{2} = 3^{2} + 1^{2} = 10 > 5 = r^{2}$. ✓ $P$ is external.

Step 2 — set up the general line through $P(3, 1)$.

Let the tangent be $y - 1 = k(x - 3)$, i.e. $kx - y + (1 - 3k) = 0$.

Step 3 — impose $d = r = \sqrt{5}$.

$$d = \frac{|k(0) - 0 + (1 - 3k)|}{\sqrt{k^{2} + 1}} = \frac{|1 - 3k|}{\sqrt{k^{2} + 1}} = \sqrt{5}$$

Squaring:

$$(1 - 3k)^{2} = 5(k^{2} + 1)$$
$$1 - 6k + 9k^{2} = 5k^{2} + 5$$
$$4k^{2} - 6k - 4 = 0 \implies 2k^{2} - 3k - 2 = 0$$
$$(2k + 1)(k - 2) = 0 \implies k = -\tfrac{1}{2}\ \text{ or }\ k = 2$$

Step 4 — write the two tangent lines.

• $k = 2$: $y - 1 = 2(x - 3) \implies 2x - y - 5 = 0$
• $k = -\tfrac{1}{2}$: $y - 1 = -\tfrac{1}{2}(x - 3) \implies x + 2y - 5 = 0$

Step 5 — check for a vertical tangent $x = 3$.

$d = \frac{|0 - 0 + (−3)|}{\sqrt{0^{2}+1^{2}}} \to$ actually use $d =$ distance from $(0,0)$ to $x=3$, which is $3$. Since $3 \ne \sqrt{5}$, no vertical tangent.

$$\boxed{2x - y - 5 = 0 \quad\text{and}\quad x + 2y - 5 = 0}$$

Try it!　自测练习

Q1. Write the standard equation of the circle with centre $(-2, 5)$ and radius $3$.

Q2. Convert $x^{2} + y^{2} + 2x - 8y + 8 = 0$ to standard form. State the centre and radius.

Q3. Does the line $x - y + 6 = 0$ intersect, touch, or miss the circle $(x - 1)^{2} + (y - 1)^{2} = 9$?

Q4. Find the tangent to $x^{2} + y^{2} = 25$ at the point $(3, -4)$.

Q5. A point $P(5, 0)$ lies outside the circle $x^{2} + y^{2} = 9$. Find the length of the tangent from $P$ to the circle.

📌 Chapter summary

Topic	Key formula / skill	核心公式/技能
8.1 Standard equation	$(x-a)^{2}+(y-b)^{2}=r^{2}$; read centre $(a,b)$ and $r=\sqrt{\text{RHS}}$ directly	标准方程，直接读取圆心和半径
8.2 General equation	Complete the square to convert; valid iff $D^{2}+E^{2}-4F>0$	配方法转化；判别式 $>0$ 才是圆
8.3 Line–circle	Compare $d$ (centre-to-line distance) with $r$: $d<r$ secant · $d=r$ tangent · $d>r$ disjoint; chord $= 2\sqrt{r^{2}-d^{2}}$	比较 $d$ 与 $r$ 的大小；弦长公式
8.4 Tangent lines	At point on circle: $(x_{0}-a)(x-a)+(y_{0}-b)(y-b)=r^{2}$; from external $P$: set $d=r$, solve quadratic in $k$; tangent length $=\sqrt{\lvert PC\rvert^{2}-r^{2}}$	切点处切线公式；外点切线解斜率

What's next → Unit 9 extends these ideas to conic sections — ellipses, hyperbolas, and parabolas are all defined by similar distance conditions, and you will use the line–conic distance technique again.