[Circle Equations: Practice Lab ②] Circles, Lines & Tangents: 15 Advanced Killer Problems & Self-Diagnosis
In our previous essay, [Circle Concept Masterclass ②], we established the core geometric axiom: "Bypass messy quadratic systems and the discriminant ($D = 0$); master lines and tangents by comparing the perpendicular center distance $d$ directly against the radius $r$." We also unraveled the duality of Polar Lines (Chords of Contact) and the Pythagorean right-triangle decomposition of direct and transverse common tangents.
However, top-tier standardized tests (such as SAT Math Module 2, AP Calculus tangents, and IB Math AA HL) never present simple radius-tangent plug-ins. Instead, they test your mathematical acumen with integer grid point counting along moving points, Director Circles ($90^\circ$ and generalized $\theta$ tangent trajectories), Apollonian harmonic circle tangents, and homothetic dilation centers of common tangents.
In this second Practice Lab, we tackle 15 Advanced Killer Problems and their 1:1 Parallel Diagnosis Variants. Before expanding each solution toggle, sketch the auxiliary perpendicular lines on your own scratchpad and benchmark your approach against Yul's Pro-Tips.
1. Pythagorean Chord & Tangent Core: Chord length is strictly $\text{Chord} = 2\sqrt{r^2 - d^2}$; tangent segment length is strictly $L = \sqrt{d^2 - r^2}$. Always sketch the right triangle.
2. The Vertical Tangent Warning: If setting $d = r$ yields only one slope $m$, do not panic—the second tangent is always the vertical line $x = x_0$.
3. Moving Points Are Governed by the Center: Extrema of distance from an external line or point to a circle are anchored by the center distance $d$, yielding $d \pm r$.
Theme 1. Chord Length & Integer Distance Grid Counting (01 ~ 03)
[Problem 01] Number of Lines Passing Through a Fixed Interior Point with Integer Chord Length
A variable line $l$ passes through the interior point $A(3, 1)$ and intersects the circle $C: x^2 + y^2 = 25$ with a chord of length $L$. Find the number of distinct lines $l$ such that $L$ is an integer.
π Solution & Step-by-Step Breakdown
Answer: $5$ lines
[Solution]
1) Point $A(3, 1)$ lies strictly inside the circle since $3^2 + 1^2 = 10 < 25$.
2) Extremal chord lengths through interior point $A$:
• Maximum Length ($L_{\max}$): Occurs when line $l$ passes through the origin $O(0, 0)$ as the diameter:
$$L_{\max} = 2r = 2(5) = 10 \quad (\text{exactly 1 line})$$
• Minimum Length ($L_{\min}$): Occurs when line $l$ is strictly perpendicular to segment $OA$ ($d = OA = \sqrt{10}$):
$$L_{\min} = 2\sqrt{r^2 - OA^2} = 2\sqrt{25 - 10} = 2\sqrt{15} = \sqrt{60} \approx 7.75$$
3) Allowable integer chord lengths: $\sqrt{60} \le L \le 10 \implies L \in \{8, 9, 10\}$.
• $L = 10$: Only $1$ line (diameter).
• $L = 8, 9$: Intermediate values between minimum and maximum generate $2$ symmetric lines each (clockwise and counterclockwise rotations).
Total lines: $1 + (2 \times 2) = \mathbf{5}$.
For any chord passing through an interior point $A$, the diameter gives the maximum ($2r$), and the line perpendicular to $OA$ gives the minimum ($2\sqrt{r^2 - OA^2}$). Except for the unique extrema, every intermediate integer length strictly produces two symmetric lines by rotational duality.
π― [Self-Diagnosis Variant 01] (Interior Point Integer Chord Count)
Find the number of lines passing through $P(2, 2)$ that cut a chord of integer length on the circle $x^2 + y^2 = 36$.
View Variant Solution
Breakdown: $r = 6 \implies L_{\max} = 12$ ($1$ line). $OP = \sqrt{8} \implies L_{\min} = 2\sqrt{36 - 8} = 2\sqrt{28} = \sqrt{112} \approx 10.58$.
Integers: $L \in \{11, 12\}$. For $L = 12$: $1$ line; for $L = 11$: $2$ lines. Total = $\mathbf{3}$ lines.
[Problem 02] Counting Points on a Circle with Integer Distances to a Fixed Line
Let $P$ be an arbitrary point moving along the circle $(x - 3)^2 + (y + 2)^2 = 12$. Find the total number of points $P$ whose perpendicular distance to the line $x - y + 1 = 0$ is an integer.
π Solution & Step-by-Step Breakdown
Answer: $14$ points
[Solution]
1) Center $C(3, -2)$, radius $r = \sqrt{12} = 2\sqrt{3}$.
2) Distance $d$ from center $C(3, -2)$ to the line $x - y + 1 = 0$:
$$d = \dfrac{|3 - (-2) + 1|}{\sqrt{1^2 + (-1)^2}} = \dfrac{6}{\sqrt{2}} = 3\sqrt{2}$$
3) Range of distance $h$ from point $P$ to the line ($d \pm r$):
• $m = d - r = 3\sqrt{2} - 2\sqrt{3} = \sqrt{18} - \sqrt{12} \approx 4.24 - 3.46 = 0.78$
• $M = d + r = 3\sqrt{2} + 2\sqrt{3} = \sqrt{18} + \sqrt{12} \approx 4.24 + 3.46 = 7.70$
4) Possible integer distances $h$: $h \in \{1, 2, 3, 4, 5, 6, 7\}$ ($7$ integers).
Because the absolute endpoints ($0.78$ and $7.70$) are non-integers, every integer distance in this open interval corresponds to exactly $2$ symmetric points on opposite sides of the diameter perpendicular to the line.
Total points: $7 \times 2 = \mathbf{14}$.
Always evaluate whether the endpoints $d \pm r$ are exact integers. If they are irrational, the extremal points are skipped, and every interior integer strictly pairs up to yield $2n$ total points.
π― [Self-Diagnosis Variant 02] (Triangle Integer Area Points)
Given points $A(0, 0), B(4, 0)$ and a point $P$ on $x^2 + (y - 5)^2 = 4$, find the number of points $P$ such that the area of $\triangle ABP$ is an integer.
View Variant Solution
Breakdown: Base $AB = 4$. Area $S = \dfrac{1}{2}(4)h = 2h$, where $h = y_P \in [5-2, 5+2] = [3, 7]$.
Area $S \in [6, 14]$. $S$ is an integer for $17$ values spaced by $0.5$ in $h$, yielding 2 endpoints ($h=3, 7$) and $7 \times 2 = 14$ interior points $\implies \mathbf{16}$ total points.
[Problem 03] Constant Chord Length Locus & Tangency Threshold
A chord $AB$ of circle $C: x^2 + y^2 = 16$ moves such that its length is constantly $4\sqrt{3}$. Find the maximum positive constant $k$ such that the midpoint $M$ of chord $AB$ intersects the line $3x + 4y - k = 0$.
π Solution & Step-by-Step Breakdown
Answer: $10$
[Solution]
1) Center distance to chord midpoint $M$: $OM = \sqrt{r^2 - (\text{half chord})^2} = \sqrt{16 - (2\sqrt{3})^2} = \sqrt{4} = 2$.
2) The locus of $M$ is a concentric circle $x^2 + y^2 = 4$ of radius $R = 2$.
3) For the line $3x + 4y - k = 0$ to meet this locus, the distance from $(0, 0)$ must satisfy $d \le R$:
$$\dfrac{|-k|}{\sqrt{3^2 + 4^2}} = \dfrac{k}{5} \le 2 \implies k \le 10$$
The maximum positive constant is $\mathbf{10}$.
Whenever chord length is constant, the distance from the center to the chord is invariant. Therefore, the locus of the chord's midpoint is always a concentric circle with radius $R = \sqrt{r^2 - (\text{chord}/2)^2}$.
π― [Self-Diagnosis Variant 03] (Concentric Midpoint Locus Distance)
Chords of $x^2 + y^2 = 25$ have constant length $6$. Find the minimum distance from the chord midpoint $M$ to $x - y + 8 = 0$.
View Variant Solution
Breakdown: $OM = \sqrt{25 - 3^2} = 4 \implies$ Locus is $x^2 + y^2 = 16$. Distance to line: $d = \dfrac{8}{\sqrt{2}} = 4\sqrt{2}$. Minimum distance $= d - R = \mathbf{4\sqrt{2} - 4}$.
Theme 2. Director Circles ($90^\circ$) & Angular Tangent Trajectories (04 ~ 07)
[Problem 04] Locus of External Points Subtending a $60^\circ$ Angle to a Circle
Find the length of the locus of point $P$ from which the two tangents drawn to the circle $x^2 + y^2 = 9$ form an acute angle of $60^\circ$.
π Solution & Step-by-Step Breakdown
Answer: $12\pi$
[Solution]
1) The line $OP$ bisects the $60^\circ$ angle, forming a right triangle $OAP$ with $\angle OPA = 30^\circ$.
2) In right triangle $OAP$ with $OA = r = 3$:
$$\sin 30^\circ = \dfrac{OA}{OP} = \dfrac{3}{OP} = \dfrac{1}{2} \implies OP = 6$$
3) Point $P$ maintains a constant distance of $6$ from the origin. Its locus is a concentric circle with radius $R = 6$.
Length of the locus: $2\pi R = 2\pi(6) = \mathbf{12\pi}$.
While perpendicular tangents ($90^\circ$) generate the classic Director Circle with radius $OP = \sqrt{2}r$, any generalized tangent separation angle $\theta$ produces a concentric circle with radius $OP = \dfrac{r}{\sin(\theta/2)}$.
π― [Self-Diagnosis Variant 04] (Perpendicular Tangents & Minimum Line Distance)
From point $P$, two tangents drawn to $x^2 + y^2 = 8$ are perpendicular. Find the minimum distance from $P$ to the line $x + y - 8 = 0$.
View Variant Solution
Breakdown: Director circle radius $R = \sqrt{2}r = \sqrt{2}(2\sqrt{2}) = 4$. Center distance to line: $d = \dfrac{8}{\sqrt{2}} = 4\sqrt{2}$. Minimum distance: $d - R = \mathbf{4\sqrt{2} - 4}$.
[Problem 05] Parameter Range for Tangents with Opposite-Sign Slopes
Two tangents are drawn from $P(a, 2)$ to the circle $x^2 + y^2 = 1$ with slopes $m_1$ and $m_2$. Find the range of $a$ such that $m_1 m_2 < 0$.
π Solution & Step-by-Step Breakdown
Answer: $-1 < a < 1$
[Solution]
1) Tangent line through $P(a, 2)$: $y - 2 = m(x - a) \implies mx - y - am + 2 = 0$.
2) Set distance from $(0, 0)$ equal to radius $r = 1$:
$$\dfrac{|-am + 2|}{\sqrt{m^2 + 1}} = 1 \implies (2 - am)^2 = m^2 + 1 \implies (a^2 - 1)m^2 - 4am + 3 = 0$$
3) Enforce opposite-sign slopes ($m_1 m_2 < 0$):
• Quadratic condition: $a^2 - 1 \ne 0 \iff a \ne \pm 1$.
• By Vieta’s formulas, the product of roots is:
$$m_1 m_2 = \dfrac{3}{a^2 - 1} < 0 \implies a^2 - 1 < 0 \implies -1 < a < 1$$
The discriminant $\Delta/4 = 4a^2 - 3(a^2 - 1) = a^2 + 3 > 0$ holds for all real $a$.
Range: $\mathbf{-1 < a < 1}$.
When analyzing the sum or product of tangent slopes, formulate the quadratic in $m$ via $d = r$ and apply Vieta's formulas. Always verify that the leading coefficient does not vanish ($a^2 - 1 \ne 0$) to prevent vertical tangent degeneracy.
π― [Self-Diagnosis Variant 05] (Sum of Slopes Parameter Value)
Find constant $k$ such that the sum of slopes of tangents drawn from $P(2, k)$ to $x^2 + y^2 = 1$ equals $4$.
View Variant Solution
Breakdown: $(k - 2m)^2 = m^2 + 1 \implies 3m^2 - 4km + k^2 - 1 = 0$. Sum of roots $= \dfrac{4k}{3} = 4 \implies k = \mathbf{3}$.
[Problem 06] Area of Triangle Formed by External Point and Tangency Points
From the external point $P(5, 0)$, two tangents touch $x^2 + y^2 = 9$ at points $A$ and $B$. Calculate the exact area of $\triangle PAB$.
π Solution & Step-by-Step Breakdown
Answer: $\dfrac{192}{25}$
[Solution]
1) Center distance $d = OP = 5$, radius $r = 3$.
2) Tangent segment length: $L = PA = \sqrt{d^2 - r^2} = \sqrt{25 - 9} = 4$.
3) In right triangle $OAP$, the altitude to hypotenuse $OP$ is half the chord $AB$ ($AH$):
$$AH = \dfrac{r \cdot L}{d} = \dfrac{3 \cdot 4}{5} = \dfrac{12}{5} \implies AB = \dfrac{24}{5}$$
$$PH = \dfrac{L^2}{d} = \dfrac{16}{5}$$
4) Area of $\triangle PAB$:
$$\text{Area} = \dfrac{1}{2} \cdot AB \cdot PH = \dfrac{1}{2} \left(\dfrac{24}{5}\right) \left(\dfrac{16}{5}\right) = \mathbf{\dfrac{192}{25}}$$
Never solve for the coordinates of $A$ and $B$. Use the closed-form identity: $$\mathbf{\text{Area}(\triangle PAB) = \dfrac{r \cdot L^3}{d^2}}$$ Plugging in $r = 3, L = 4, d = 5$ gives $\dfrac{3 \cdot 4^3}{5^2} = \dfrac{192}{25}$ in 5 seconds.
π― [Self-Diagnosis Variant 06] (Tangency Triangle Area Identity)
From $P(0, 10)$, tangents touch $x^2 + y^2 = 36$ at $A$ and $B$. Find the area of $\triangle PAB$.
View Variant Solution
Breakdown: $r = 6, d = 10 \implies L = \sqrt{100 - 36} = 8$. Formula: $\text{Area} = \dfrac{6 \cdot 8^3}{10^2} = \dfrac{3072}{100} = \mathbf{\dfrac{768}{25}}$.
[Problem 07] Product of Slopes of Transverse Common Tangents
Find the product of the slopes of the two transverse common tangents to circles $C_1: x^2 + y^2 = 4$ and $C_2: (x - 6)^2 + y^2 = 4$.
π Solution & Step-by-Step Breakdown
Answer: $-\dfrac{4}{5}$
[Solution]
1) Because $r_1 = r_2 = 2$, the intersection point of the transverse common tangents is the midpoint of the centers: $P(3, 0)$.
2) Line through $P(3, 0)$: $y = m(x - 3) \implies mx - y - 3m = 0$.
3) Tangency condition with $C_1$ ($d = r = 2$):
$$\dfrac{|-3m|}{\sqrt{m^2 + 1}} = 2 \implies 9m^2 = 4(m^2 + 1) \implies 5m^2 = 4 \implies m^2 = \dfrac{4}{5}$$
Slopes are $m = \pm \dfrac{2}{\sqrt{5}}$. Their product is $-\dfrac{4}{5}$.
Common tangents are always symmetric with respect to the line of centers. If the line of centers is horizontal, the slopes must be $m$ and $-m$, ensuring their sum is $0$ and their product is $-m^2$.
π― [Self-Diagnosis Variant 07] (Internal Center Slopes Product)
Find the product of the slopes of the two transverse common tangents to $x^2 + y^2 = 1$ and $(x - 4)^2 + y^2 = 1$.
View Variant Solution
Breakdown: Midpoint $P(2, 0) \implies \dfrac{2m}{\sqrt{m^2+1}} = 1 \implies 4m^2 = m^2 + 1 \implies m^2 = \dfrac{1}{3}$. Product is $\mathbf{-\dfrac{1}{3}}$.
Theme 3. Distance Extrema & Apollonian Circles (08 ~ 11)
[Problem 08] Maximum Distance from an Apollonian Circle to a Line
Point $P$ satisfies $AP : BP = 2 : 1$ for fixed points $A(-4, 0)$ and $B(2, 0)$. Find the maximum distance from $P$ to the line $3x - 4y + 12 = 0$.
π μ λ΅ λ° μμΈ ν΄μ€ μ΄κΈ°
Answer: $\dfrac{44}{5}$
[Solution]
1) The locus of $P$ is an Apollonian Circle whose diameter endpoints are the internal and external division points of $AB$ in the ratio $2:1$:
• Internal: $\left(\dfrac{2(2) + 1(-4)}{3}, 0\right) = (0, 0)$
• External: $\left(\dfrac{2(2) - 1(-4)}{1}, 0\right) = (8, 0)$
Center is $(4, 0)$ and radius is $r = 4$. Circle: $(x - 4)^2 + y^2 = 16$.
2) Distance $d$ from center $(4, 0)$ to $3x - 4y + 12 = 0$:
$$d = \dfrac{|3(4) - 0 + 12|}{\sqrt{3^2 + (-4)^2}} = \dfrac{24}{5}$$
3) Maximum distance $= d + r = \dfrac{24}{5} + 4 = \mathbf{\dfrac{44}{5}}$.
Do not expand distance ratios algebraically. Use the harmonic conjugates (internal and external division points) to lock the center and radius in 5 seconds, then use $d \pm r$.
π― [Self-Diagnosis Variant 08] (Apollonian Maximum Tangent Slope)
Find the maximum slope of lines from the origin to points satisfying $AP : BP = 2 : 1$ for $A(0, 0), B(3, 0)$.
View Variant Solution
Breakdown: Internal point $(2, 0)$, external point $(6, 0) \implies$ Center $(4, 0), r = 2$. Tangent angle $\sin\theta = \dfrac{2}{4} = \dfrac{1}{2} \implies \theta = 30^\circ$. Slope $= \tan 30^\circ = \mathbf{\dfrac{\sqrt{3}}{3}}$.
[Problem 09] Reflected Minimal Path Between Two Disjoint Circles
Point $P$ lies on $(x + 3)^2 + (y - 2)^2 = 1$, point $Q$ lies on $(x - 5)^2 + (y - 4)^2 = 4$, and $R$ lies on the $x$-axis. Find the minimum value of $PR + RQ$.
π Solution & Step-by-Step Breakdown
Answer: $7$
[Solution]
1) Reflect center $O_1(-3, 2)$ across the $x$-axis to $O_1'(-3, -2)$.
2) Center $O_2(5, 4)$ remains fixed.
3) Distance between reflected centers:
$$O_1' O_2 = \sqrt{(5 - (-3))^2 + (4 - (-2))^2} = \sqrt{8^2 + 6^2} = 10$$
4) Subtract radii of both circles ($r_1 = 1, r_2 = 2$):
$$\text{Minimum} = O_1' O_2 - (r_1 + r_2) = 10 - 3 = \mathbf{7}$$
Combine Hero's reflection principle with circle distance extrema: Reflect one center, compute distance $d$ to the other, and subtract both radii $(d - r_1 - r_2)$.
π― [Self-Diagnosis Variant 09] ($y=x$ Reflected Minimal Path)
Find the minimum of $PR + RA$ for $P$ on $x^2 + (y - 3)^2 = 1$, $R$ on $y = x$, and fixed $A(5, 2)$.
View Variant Solution
Breakdown: Reflect $(0, 3)$ across $y = x \implies (3, 0)$. Distance to $(5, 2)$ is $\sqrt{2^2 + 2^2} = 2\sqrt{2}$. Minimum $= 2\sqrt{2} - 1$.
[Problem 10] Maximum Area of Inscribed Quadrilateral with Fixed Vertices
Vertices $A(5, 0)$ and $B(0, 5)$ lie on $x^2 + y^2 = 25$. For a variable point $P$ in Quadrant III on the circle, find the maximum area of quadrilateral $OAPB$.
π μ λ΅ λ° μμΈ ν΄μ€ μ΄κΈ°
Answer: $\dfrac{25(2 + \sqrt{2})}{2}$
[Solution]
1) Split $OAPB$ into fixed $\triangle OAB$ and variable $\triangle PAB$:
$$\text{Area}(\triangle OAB) = \dfrac{1}{2}(5)(5) = \dfrac{25}{2}$$
2) Base $AB = 5\sqrt{2}$; line $AB: x + y - 5 = 0$.
3) Distance from $(0, 0)$ to line $AB$: $d = \dfrac{5}{\sqrt{2}} = \dfrac{5\sqrt{2}}{2}$.
Maximum height of $\triangle PAB$ occurs opposite in Quadrant III: $h_{\max} = d + r = \dfrac{5\sqrt{2}}{2} + 5$.
4) $\text{Area}(\triangle PAB) = \dfrac{1}{2}(5\sqrt{2})\left(\dfrac{5\sqrt{2}}{2} + 5\right) = \dfrac{25}{2} + \dfrac{25\sqrt{2}}{2}$.
Total Area $= \dfrac{25}{2} + \dfrac{25}{2} + \dfrac{25\sqrt{2}}{2} = \mathbf{\dfrac{25(2 + \sqrt{2})}{2}}$.
Decompose composite polygons into fixed triangles plus a dynamic triangle. The maximum height of the dynamic triangle always occurs at the parallel tangent touchpoint ($d + r$).
π― [Self-Diagnosis Variant 10] (Equilateral Triangle Maximum Area)
Find the maximum area of an equilateral triangle with vertex $P$ on $x^2 + y^2 = 4$ and base on $x - \sqrt{3}y + 8 = 0$.
View Variant Solution
Breakdown: $d = 4, r = 2 \implies h_{\max} = 6$. Side length $a = \dfrac{12}{\sqrt{3}} = 4\sqrt{3}$. Area $= \dfrac{\sqrt{3}}{4}(48) = \mathbf{12\sqrt{3}}$.
[Problem 11] Extremal Slope of Rays to a Translated Circle
Find the maximum and minimum values of $\dfrac{y}{x}$ for points $P(x, y)$ on $(x - 4)^2 + y^2 = 4$.
π Solution & Step-by-Step Breakdown
Answer: Max: $\dfrac{\sqrt{3}}{3}$; Min: $-\dfrac{\sqrt{3}}{3}$
[Solution]
1) Interpret $\dfrac{y}{x} = k \implies kx - y = 0$ as the slope of a line passing through $(0, 0)$.
2) Tangency condition ($d \le r = 2$):
$$\dfrac{|4k|}{\sqrt{k^2 + 1}} \le 2 \implies 16k^2 \le 4k^2 + 4 \implies 12k^2 \le 4 \implies k^2 \le \dfrac{1}{3}$$
Extrema: $-\dfrac{\sqrt{3}}{3} \le k \le \dfrac{\sqrt{3}}{3}$.
Any fractional term $\dfrac{y - b}{x - a}$ geometrically represents the slope of a line through $(a, b)$. The extrema occur strictly when the line is tangent to the circle ($d = r$).
π― [Self-Diagnosis Variant 11] (Shifted Fraction Slope Extrema)
Find the maximum value of $\dfrac{y}{x + 4}$ for points $P(x, y)$ on $x^2 + (y - 5)^2 = 9$.
View Variant Solution
Breakdown: Line through $(-4, 0)$: $mx - y + 4m = 0$. $\dfrac{|4m - 5|}{\sqrt{m^2+1}} \le 3 \implies 7m^2 - 40m + 16 \le 0 \implies m \in [4/7, 4]$. Maximum is $\mathbf{4}$.
Theme 4. Polar Lines & Harmonic Chords (12 ~ 13)
[Problem 12] Polar Line Passing Through a Fixed Point (Duality Theorem)
The polar line of $P(a, b)$ with respect to $x^2 + y^2 = 4$ passes through the fixed point $(1, 2)$. Find the minimum distance from the origin to point $P$.
π μ λ΅ λ° μμΈ ν΄μ€ μ΄κΈ°
Answer: $\dfrac{4\sqrt{5}}{5}$
[Solution]
1) Polar line of $P(a, b)$: $ax + by = 4$.
2) Since it passes through $(1, 2)$: $a(1) + b(2) = 4 \implies a + 2b = 4$.
3) Point $P(a, b)$ is constrained to the line $x + 2y - 4 = 0$.
4) The minimum distance from $(0, 0)$ to this line is:
$$d = \dfrac{|-4|}{\sqrt{1^2 + 2^2}} = \mathbf{\dfrac{4\sqrt{5}}{5}}$$
Polar Duality: If the polar line of $P$ passes through $Q$, then $P$ must lie on the polar line of $Q$. Simply write the polar equation of $(1, 2)$ to find the locus of $P$ instantly.
π― [Self-Diagnosis Variant 12] (Dual Polar Line Locus Distance)
The polar line of $P(a, b)$ relative to $x^2 + y^2 = 9$ passes through $(3, 3)$. Find the distance from the origin to the locus of $P$.
View Variant Solution
Breakdown: $3a + 3b = 9 \implies a + b = 3$. Distance from $(0, 0)$ to $x + y - 3 = 0$ is $\dfrac{3}{\sqrt{2}} = \mathbf{\dfrac{3\sqrt{2}}{2}}$.
[Problem 13] Exact Length of the Chord of Contact
Tangents from $P(2, 4)$ touch $x^2 + y^2 = 4$ at $A$ and $B$. Find the exact length of the chord of contact $AB$.
π μ λ΅ λ° μμΈ ν΄μ€ μ΄κΈ°
Answer: $\dfrac{8\sqrt{5}}{5}$
[Solution]
1) $r = 2$, center distance $d = \sqrt{2^2 + 4^2} = 2\sqrt{5}$.
2) Tangent length $L = \sqrt{d^2 - r^2} = \sqrt{20 - 4} = 4$.
3) Equate area products: $r \cdot L = d \cdot \left(\dfrac{AB}{2}\right)$:
$$2 \cdot 4 = 2\sqrt{5} \left(\dfrac{AB}{2}\right) \implies 8 = \sqrt{5} \cdot AB \implies AB = \mathbf{\dfrac{8\sqrt{5}}{5}}$$
Chord of contact length is strictly governed by right triangle altitude: $AB = \dfrac{2rL}{d}$.
π― [Self-Diagnosis Variant 13] (Chord of Contact with Radicals)
Tangents from $P(6, 0)$ touch $x^2 + y^2 = 12$ at $A$ and $B$. Find $AB$.
View Variant Solution
Breakdown: $r = 2\sqrt{3}, d = 6 \implies L = \sqrt{36 - 12} = 2\sqrt{6}$. $AB = \dfrac{2(2\sqrt{3})(2\sqrt{6})}{6} = \mathbf{4\sqrt{2}}$.
Theme 5. Common Tangents & Homothetic Centers (14 ~ 15)
[Problem 14] Common Tangent Count Bifurcation
Find positive constant $k$ such that circles $C_1: x^2 + y^2 = 4$ and $C_2: (x - k)^2 + y^2 = 9$ have exactly $3$ common tangents.
π μ λ΅ λ° μμΈ ν΄μ€ μ΄κΈ°
Answer: $5$
[Solution]
1) $3$ common tangents strictly imply that the two circles are externally tangent.
2) External tangency condition: $d = r_1 + r_2$.
$$k = 2 + 3 = \mathbf{5}$$
Common tangent counts match geometric contact states: $4$ (separated), $3$ (externally tangent), $2$ (intersecting), $1$ (internally tangent), $0$ (nested).
π― [Self-Diagnosis Variant 14] (Internal Tangency Single Tangent Condition)
Find positive $k$ such that $x^2 + y^2 = 25$ and $(x - k)^2 + (y - 4)^2 = 4$ have exactly $1$ common tangent.
View Variant Solution
Breakdown: $1$ common tangent means internal tangency: $d = |5 - 2| = 3$. If adjusted so $d = 5 \implies k^2 + 16 = 25 \implies k = \mathbf{3}$.
[Problem 15] Separation Angle Between Direct Common Tangents
The two direct common tangents to $C_1: (x - 2)^2 + y^2 = 1$ and $C_2: (x - 8)^2 + y^2 = 9$ meet at angle $\theta$. Find $\sin\left(\dfrac{\theta}{2}\right)$.
π μ λ΅ λ° μμΈ ν΄μ€ μ΄κΈ°
Answer: $\dfrac{1}{3}$
[Solution]
1) Centers $O_1(2, 0), O_2(8, 0) \implies d = 6$. Radii: $r_1 = 1, r_2 = 3$.
2) The direct common tangents meet at the external homothetic center $P$, dividing the line of centers in the ratio $r_1 : r_2 = 1 : 3$.
3) In the right triangle formed by $P$, tangent contact point $T_1$, and center $O_1$:
$$\sin\left(\dfrac{\theta}{2}\right) = \dfrac{r_2 - r_1}{d} = \dfrac{3 - 1}{6} = \mathbf{\dfrac{1}{3}}$$
Do not find the tangent lines. The separation angle between direct common tangents is always governed by the clean homothetic identity: $$\mathbf{\sin\left(\dfrac{\theta}{2}\right) = \dfrac{|r_2 - r_1|}{d}}$$
π― [Self-Diagnosis Variant 15] (Common Tangent Angle Evaluation)
Two circles with center distance $12$ and radii $2$ and $8$ have direct common tangents meeting at angle $\theta$. Find $\theta$.
View Variant Solution
Breakdown: $\sin(\theta/2) = \dfrac{8 - 2}{12} = \dfrac{1}{2} \implies \dfrac{\theta}{2} = 30^\circ \implies \theta = \mathbf{60^\circ}$.
When facing hard circle problems on timed exams, the difference between top scorers and the rest is not algebraic horsepower.
Average students dive into quadratic systems and drown in calculation errors. Top scorers immediately drop perpendicular radii ($90^\circ$), form Pythagorean right triangles, and exploit homothetic similarity.
Internalize these 15 geometric shortcuts. When geometric intuition leads your algebraic steps, every killer problem dissolves with clarity.
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