9.3 Positional Relationships Between Lines and Circles

9.3 Positional Relationships Between Lines and Circles

The positional relationship between a straight line and a circle is a fundamental topic in coordinate geometry. In Gaokao Mathematics, line-circle problems test geometric construction, chord calculations, tangency conditions, chord of contact equations, midpoint chord properties, and optimization of distance and area metrics.

---

1. Geometric and Algebraic Criteria for Line-Circle Relationships

Consider line $l: Ax + By + C = 0$ ($A^2 + B^2 \neq 0$) and circle $C: (x - a)^2 + (y - b)^2 = r^2$ ($r > 0$).

Two Determination Methods

1. Geometric Method (几何法): Compute the perpendicular distance $d$ from center $C(a, b)$ to line $l$:

   $$d = \frac{|Aa + Bb + C|}{\sqrt{A^2 + B^2}}$$
   • Secant Line (相交): $d < r \iff$ Line $l$ intersects circle $C$ at two distinct points.
   • Tangent Line (相切): $d = r \iff$ Line $l$ is tangent to circle $C$ at exactly one point.
   • Disjoint Line (相离): $d > r \iff$ Line $l$ and circle $C$ have no common real points.

2. Algebraic Method (代数法): Substitute line equation $y = kx + m$ (or $x = cy + n$) into circle equation $(x - a)^2 + (y - b)^2 = r^2$ to obtain a single-variable quadratic equation $A x^2 + B x + C = 0$. Calculate discriminant $\Delta = B^2 - 4AC$:

   • Secant Line: $\Delta > 0 \iff 2$ distinct real intersection points.
   • Tangent Line: $\Delta = 0 \iff 1$ real intersection point (tangency).
   • Disjoint Line: $\Delta < 0 \iff 0$ real intersection points.

Gaokao Strategy: Always prefer the Geometric Method when analyzing line-circle relationships. The geometric distance formula avoids expanding quadratic terms and eliminates heavy algebraic computations.

---

2. Secant Lines and Chord Length Formulas

When a line $l$ intersects a circle $C(r)$ at points $A$ and $B$, segment $AB$ is a chord.

The Geometric Right-Triangle Chord Formula

Let $M$ be the midpoint of chord $AB$. By Euclidean circle geometry, radius $CM \perp AB$. In right triangle $\triangle CAM$:

$$|AM|^2 + |CM|^2 = |CA|^2 \implies \left(\frac{L_{AB}}{2}\right)^2 + d^2 = r^2$$ $$L_{AB} = 2 \sqrt{r^2 - d^2}$$

where $d = \frac{|Aa + Bb + C|}{\sqrt{A^2 + B^2}}$ is the center-to-line distance.

Algebraic Vieta Chord Length Formula

If line $l: y = kx + m$ intersects circle $C$ at $A(x_1, y_1)$ and $B(x_2, y_2)$, substituting $y = kx + m$ yields quadratic equation $ax^2 + bx + c = 0$ with discriminant $\Delta > 0$. By Vieta's formulas ($x_1 + x_2 = -\frac{b}{a}, x_1 x_2 = \frac{c}{a}$):

$$L_{AB} = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} = \sqrt{(1 + k^2)(x_1 - x_2)^2} = \sqrt{1 + k^2} \cdot \sqrt{(x_1 + x_2)^2 - 4x_1 x_2} = \sqrt{1 + k^2} \frac{\sqrt{\Delta}}{|a|}$$

Midpoint of Chord & Point-Difference Method (点差法)

Let $M(x_0, y_0)$ be the midpoint of chord $AB$ on circle $(x - a)^2 + (y - b)^2 = r^2$:

• Since $A(x_1, y_1)$ and $B(x_2, y_2)$ lie on the circle: $$(x_1 - a)^2 + (y_1 - b)^2 = r^2$$ $$(x_2 - a)^2 + (y_2 - b)^2 = r^2$$
• Subtracting the two equations: $$(x_1 - x_2)(x_1 + x_2 - 2a) + (y_1 - y_2)(y_1 + y_2 - 2b) = 0$$
• Substituting $x_1 + x_2 = 2x_0$ and $y_1 + y_2 = 2y_0$: $$(x_1 - x_2)(2x_0 - 2a) + (y_1 - y_2)(2y_0 - 2b) = 0 \implies \frac{y_1 - y_2}{x_1 - x_2} \cdot \frac{y_0 - b}{x_0 - a} = -1$$ $$k_{AB} \cdot k_{CM} = -1$$

Conclusion: The chord $AB$ is perpendicular to the ray connecting the center $C(a, b)$ to the chord's midpoint $M(x_0, y_0)$.

---

3. Tangent Lines, Tangent Segments, and Chords of Contact

Tangent Line Equations

1. Point $P_0(x_0, y_0)$ ON circle $x^2 + y^2 = r^2$: The tangent line equation is: $$x_0 x + y_0 y = r^2$$
2. Point $P_0(x_0, y_0)$ ON circle $(x - a)^2 + (y - b)^2 = r^2$: The tangent line equation is: $$(x_0 - a)(x - a) + (y_0 - b)(y - b) = r^2$$

Tangents from an External Point $P_0(x_0, y_0)$

From an external point $P_0(x_0, y_0)$ outside circle $C(a, b; r)$ ($|P_0 C| > r$), exactly two tangent lines can be drawn to the circle.

• Method: Set slope of tangent line as $k$: $y - y_0 = k(x - x_0) \implies kx - y + (y_0 - kx_0) = 0$. Apply geometric distance condition $d = r$: $$\frac{|ka - b + y_0 - kx_0|}{\sqrt{k^2 + 1}} = r$$ Squaring yields a quadratic equation in $k$.
• Caution: If the quadratic in $k$ yields only one real solution, the second tangent line is vertical ($x = x_0$, undefined slope). Always test vertical line $x = x_0$ separately!

Tangent Segment Length

For external point $P_0(x_0, y_0)$ and tangency point $T$, right triangle $\triangle P_0 T C$ yields:

$$l_{tangent} = |P_0 T| = \sqrt{|P_0 C|^2 - r^2}$$

For general circle equation $x^2 + y^2 + Dx + Ey + F = 0$:

$$l_{tangent} = \sqrt{x_0^2 + y_0^2 + D x_0 + E y_0 + F}$$

Chord of Contact (切点弦)

If two tangents drawn from external point $P_0(x_0, y_0)$ touch circle $x^2 + y^2 = r^2$ at $T_1(x_1, y_1)$ and $T_2(x_2, y_2)$, line $T_1 T_2$ is called the chord of contact (切点弦).

• Equation of Chord of Contact: $$x_0 x + y_0 y = r^2$$

Proof: Tangent at $T_1$ is $x_1 x + y_1 y = r^2$. Since $P_0(x_0, y_0)$ lies on it, $x_1 x_0 + y_1 y_0 = r^2 \implies x_0 x_1 + y_0 y_1 = r^2$. Similarly for $T_2$, $x_0 x_2 + y_0 y_2 = r^2$. Thus both $T_1$ and $T_2$ satisfy linear equation $x_0 x + y_0 y = r^2$.

---

4. Distance Extrema and Area Optimization

Minimum and Maximum Distance from Circle to Line

Let line $l$ be disjoint from circle $C(a, b; r)$ ($d_{center} > r$). For any point $P$ on circle $C$ and point $Q$ on line $l$:

• Minimum distance between circle and line: $$d_{\min} = d_{center} - r$$
• Maximum distance from a point on circle to line $l$: $$d_{\max} = d_{center} + r$$

where $d_{center} = \frac{|Aa + Bb + C|}{\sqrt{A^2 + B^2}}$.

Optimization of Triangle Area $\triangle PAB$

If chord $AB$ is fixed on circle $C$, the area of triangle $\triangle PAB$ with $P$ moving on circle $C$ is $S_{\triangle PAB} = \frac{1}{2} |AB| \cdot h_P$. The height $h_P$ reaches its maximum $d_{center} + r$ when $P$ is at the furthest point from secant line $AB$.