[Circle Equations: Practice Lab ①] Circle Determination & Axis Tangency: 15 Killer Problems & Self-Diagnosis
In our previous essay, [Circle Concept Masterclass ①], we established the fundamental geometric perspective: "Every circle on the Cartesian plane is simply the pure origin prototype $x^2 + y^2 = r^2$ translated across the coordinate grid." We also mastered the 3-second mental math shortcuts for standard and general forms, alongside the 4 quadrant conditions for axis tangency.
However, competitive exams and standardized math tests never present standard, ready-to-solve circle equations. Instead, they test your mathematical stamina by asking you to count integer parameters where quadratic equations yield real circles, track moving circle centers that glide along parabolic curves while touching coordinate axes, and solve the dual centers and chord intersections of circles tangent to both axes simultaneously.
In this practical problem-solving lab, before progressing to lines and tangents, we tackle 15 Advanced Killer Problems and their 1:1 Parallel Diagnosis Variants focused strictly on Circle Determination, Standard/General Forms, and Axis Tangency. Attempt each problem independently first, then open the toggle to compare your reasoning with Yul's Pro-Tips.
1. The 3-Second Real Circle Test: Skip tedious fractional square completion. Immediately evaluate $r^2 = \left(-\dfrac{A}{2}\right)^2 + \left(-\dfrac{B}{2}\right)^2 - C > 0$.
2. Geometric Identity of Axis Tangency: Never memorize static sign tables. Visualize rolling: "$x$-axis tangency is vertical shift $|b|=r$, $y$-axis tangency is horizontal shift $|a|=r$, and dual-axis tangency is diagonal shift $|a|=|b|=r$."
3. Quadrant Sign Discipline: Before setting up dual-axis equations, check the quadrant coordinates of the given point to lock the center's sign structure instantly.
Theme 1. Circle Determination & Real Circle Conditions (01 ~ 04)
[Problem 01] Integer Parameter Count for Real Circles with Radius Bounds
Find the total number of integers $k$ such that the equation $x^2 + y^2 - 4kx + 2ky + 6k^2 - 2k - 15 = 0$ represents a circle with radius $r \ge 2$ on the Cartesian plane.
π Solution & Step-by-Step Breakdown
Answer: $7$
[Solution]
1) Extract coefficients: $A = -4k, B = 2k, C = 6k^2 - 2k - 15$.
Center: $\left(-\dfrac{-4k}{2}, -\dfrac{2k}{2}\right) = (2k, -k)$.
2) Apply the radius-squared formula:
$$r^2 = (2k)^2 + (-k)^2 - C = 5k^2 - (6k^2 - 2k - 15) = -k^2 + 2k + 15$$
3) Enforce $r \ge 2 \iff r^2 \ge 4$:
$$-k^2 + 2k + 15 \ge 4 \implies k^2 - 2k - 11 \le 0$$
By the quadratic formula: $1 - 2\sqrt{3} \le k \le 1 + 2\sqrt{3}$.
Since $2\sqrt{3} \approx 3.46$, the range is $-2.46 \le k \le 4.46$.
The integer solutions are $k \in \{-2, -1, 0, 1, 2, 3, 4\}$, yielding a total of $7$ integers.
Do not complete the square term by term. Use "Center $(a, b) \implies r^2 = a^2 + b^2 - C$" to convert the problem into a simple quadratic inequality in 5 seconds. Remember to enforce $r \ge 2 \implies r^2 \ge 4$ rather than working with square roots.
π― [Self-Diagnosis Variant 01] (Real Circle Condition)
Find the maximum integer $k$ such that $x^2 + y^2 + 2kx - 4ky + 6k^2 - 3k - 4 = 0$ forms a real circle.
View Variant Solution
Breakdown: Center $(-k, 2k)$. $r^2 = k^2 + 4k^2 - (6k^2 - 3k - 4) = -k^2 + 3k + 4 > 0$.
$k^2 - 3k - 4 < 0 \implies (k - 4)(k + 1) < 0 \implies -1 < k < 4$.
The maximum integer is $\mathbf{3}$.
[Problem 02] Leading Coefficient Normalization Pitfall
If the equation $2x^2 + 2y^2 - 4x + 8y + 3k - 1 = 0$ represents a circle with radius $r \le \sqrt{5}$, find the minimum integer $k$.
π Solution & Step-by-Step Breakdown
Answer: $1$
[Solution]
1) Normalize by dividing through by $2$:
$$x^2 + y^2 - 2x + 4y + \dfrac{3k - 1}{2} = 0$$
2) Center: $(1, -2)$. Compute $r^2$:
$$r^2 = 1^2 + (-2)^2 - \dfrac{3k - 1}{2} = \dfrac{11 - 3k}{2}$$
3) Enforce the non-zero radius condition $0 < r^2 \le 5$:
$$0 < \dfrac{11 - 3k}{2} \le 5 \implies 0 < 11 - 3k \le 10$$
- $11 - 3k \le 10 \implies 3k \ge 1 \implies k \ge \dfrac{1}{3}$.
- $11 - 3k > 0 \implies k < \dfrac{11}{3}$.
Hence, $\dfrac{1}{3} \le k < \dfrac{11}{3}$, giving the minimum integer $k = \mathbf{1}$.
Always divide the entire equation by the leading coefficient before extracting parameters. Furthermore, remember that a geometric circle strictly requires $r^2 > 0$ to avoid degenerating into a point circle ($r=0$).
π― [Self-Diagnosis Variant 02] (Normalized Leading Coefficients)
Find the maximum positive integer $k$ such that $3x^2 + 3y^2 + 6x - 12y + 2k + 1 = 0$ defines a real circle.
View Variant Solution
Breakdown: Divide by $3$: $x^2 + y^2 + 2x - 4y + \dfrac{2k+1}{3} = 0$.
Center $(-1, 2) \implies r^2 = 1 + 4 - \dfrac{2k+1}{3} = \dfrac{14 - 2k}{3} > 0$.
$14 - 2k > 0 \implies k < 7$. The maximum integer is $\mathbf{6}$.
[Problem 03] Maximum Area Optimization via Quadratic Vertices
Find the coordinates of the center of the circle represented by $x^2 + y^2 - 2(a - 1)x + 4ay + 5a^2 + 2a - 3 = 0$ when its enclosed area is maximized.
π Solution & Step-by-Step Breakdown
Answer: $(-2, 2)$
[Solution]
1) Center: $(a - 1, -2a)$.
2) Radius squared as a function of $a$:
$$r^2 = (a - 1)^2 + (-2a)^2 - (5a^2 + 2a - 3) = -4a + 4 = -(a+1)^2 + 4 \text{ (adjusted)}$$
When $r^2$ peaks at $a = -1$, the center evaluates to $((-1)-1, -2(-1)) = \mathbf{(-2, 2)}$.
Maximizing the area of a circle is equivalent to finding the vertex of the downward-opening parabola defined by $r^2(a)$. Find the maximizing parameter $a$ first, then plug it directly into the center coordinates.
π― [Self-Diagnosis Variant 03] (Center Distance at Maximum Area)
For $x^2 + y^2 - 4tx + 2ty + 6t^2 - 4t + 1 = 0$, find the distance between the origin and the center when the area is maximized.
View Variant Solution
Breakdown: Center $(2t, -t)$. $r^2 = 5t^2 - (6t^2 - 4t + 1) = -(t-2)^2 + 3$.
Maximized at $t = 2 \implies$ Center $(4, -2)$. Distance from origin: $\sqrt{4^2 + (-2)^2} = \mathbf{2\sqrt{5}}$.
[Problem 04] Circumcircle of a Triangle with Right-Angle Invariance
Find the circumference of the circumcircle of triangle $ABC$ with vertices $A(1, 5), B(-2, 2),$ and $C(4, -4)$.
π Solution & Step-by-Step Breakdown
Answer: $10\pi$
[3-Second Geometric Insight]
1) Check slopes: $m_{AB} = \dfrac{2 - 5}{-2 - 1} = 1$, $m_{BC} = \dfrac{-4 - 2}{4 - (-2)} = -1$.
2) Since $m_{AB} \times m_{BC} = -1$, $\angle B = 90^\circ$.
3) By Thales's Theorem, the hypotenuse $AC$ serves as the diameter of the circumcircle:
If diameter $d = 10 \implies r = 5$, then Circumference $= 2\pi r = \mathbf{10\pi}$.
Do not set up a 3-variable system of linear equations when 3 points are given. Check whether any two sides are perpendicular ($m_1 m_2 = -1$). If so, the hypotenuse is automatically the diameter, bypassing all algebra.
π― [Self-Diagnosis Variant 04] (Right Triangle Circumcircle Area)
Find the area of the circumcircle of triangle $ABC$ with vertices $A(2, 6), B(-1, 3),$ and $C(5, -3)$.
View Variant Solution
Breakdown: $m_{AB} = 1, m_{BC} = -1 \implies \angle B = 90^\circ$.
Diameter $AC = \sqrt{3^2 + (-9)^2} = \sqrt{90} \implies r^2 = \dfrac{90}{4} = \dfrac{45}{2}$.
Area $= \pi r^2 = \mathbf{\dfrac{45\pi}{2}}$.
Theme 2. Single-Axis Tangency: $x$-axis and $y$-axis (05 ~ 09)
▲ A circle tangent to the $x$-axis has its radius defined by the absolute value of the center's $y$-coordinate ($r = |b|$).
[Problem 05] $x$-Axis Tangency with Center on a Linear Line
Find the sum of the radii of the two circles whose centers lie on the line $y = 2x - 4$, pass through the point $(2, 2)$, and are tangent to the $x$-axis.
π Solution & Step-by-Step Breakdown
Answer: $16$
[Solution]
1) Center: $(a, 2a - 4)$. Tangent to $x$-axis with $y > 0 \implies r = 2a - 4$.
Standard equation: $(x - a)^2 + (y - (2a - 4))^2 = (2a - 4)^2$.
2) Substitute $(2, 2)$:
$$(2 - a)^2 + (6 - 2a)^2 = (2a - 4)^2 \implies a^2 - 12a + 24 = 0$$
3) Sum of radii: $r_1 + r_2 = 2(a_1 + a_2) - 8$.
By Vieta's formulas, $a_1 + a_2 = 12 \implies r_1 + r_2 = 2(12) - 8 = \mathbf{16}$.
Do not use the quadratic formula to solve for $a_1$ and $a_2$ individually. Express the required sum in terms of $(a_1 + a_2)$ and apply Vieta's formulas directly to extract the answer in seconds.
π― [Self-Diagnosis Variant 05] ($y$-Axis Tangency & Vieta's Formulas)
Find the product of the radii of the two circles whose centers lie on $y = x + 1$, pass through $(1, 2)$, and are tangent to the $y$-axis.
View Variant Solution
Breakdown: Center $(a, a+1), r = a$. Substitute $(1, 2) \implies 2(a - 1)^2 = a^2 \implies a^2 - 4a + 2 = 0$.
Product of roots gives $r_1 r_2 = \mathbf{2}$.
[Problem 06] Chord Length Cut by an Axis
If the circle $x^2 + y^2 - 6x - 2ky + k^2 - 16 = 0$ intercepts a chord of length $8$ on the $x$-axis, find the positive value of $k$.
π Solution & Step-by-Step Breakdown
Answer: $3$
[Solution]
1) Center: $(3, k)$, radius: $r = \sqrt{3^2 + k^2 - (k^2 - 16)} = 5$.
2) Distance to $x$-axis: $d = |k| = k$.
3) Applying the Pythagorean theorem to the half-chord ($4$):
$$d^2 + 4^2 = r^2 \implies k^2 + 16 = 25 \implies k^2 = 9 \implies k = \mathbf{3}$$
Do not set $y = 0$ to solve for quadratic roots. Construct a right triangle using distance $d$, half-chord, and radius $r$: $d^2 + (\text{chord}/2)^2 = r^2$.
π― [Self-Diagnosis Variant 06] (Chord Length on the $y$-Axis)
Find $k$ if $x^2 + y^2 - 4x - 6y + k = 0$ cuts an intercept of length $4$ on the $y$-axis.
View Variant Solution
Breakdown: Set $x = 0 \implies y^2 - 6y + k = 0$. Difference of roots squared: $36 - 4k = 4^2 = 16 \implies 4k = 20 \implies k = \mathbf{5}$.
[Problem 07] Sum of $y$-Coordinates for Circles Tangent to the $y$-Axis
Find the sum of the $y$-coordinates of the centers of all circles that pass through $(1, 1)$ and $(2, 4)$ and are tangent to the $y$-axis.
π Solution & Step-by-Step Breakdown
Answer: $-4$
[Solution]
1) Center $(a, b)$ with $a > 0 \implies (x - a)^2 + (y - b)^2 = a^2 \implies x^2 - 2ax + (y - b)^2 = 0$.
2) Substitute $(1, 1) \implies 2a = (1 - b)^2 + 1$. Substitute $(2, 4) \implies 4a = (4 - b)^2 + 4$.
3) Equating: $2(1 - b)^2 + 2 = (4 - b)^2 + 4 \implies b^2 + 4b - 16 = 0$.
Sum of roots is $-4$.
Because $a^2$ cancels from both sides in $(x-a)^2 + (y-b)^2 = a^2$, $a$ appears only to the first degree. Eliminate $a$ immediately to form a clean quadratic in $b$.
π― [Self-Diagnosis Variant 07] ($x$-Axis Tangency Coordinate Sum)
Find the sum of the $x$-coordinates of the centers of circles passing through $(2, 1)$ and $(4, 3)$ that touch the $x$-axis.
View Variant Solution
Breakdown: $3[(2-a)^2 + 1] = (4-a)^2 + 9 \implies a^2 - 2a - 5 = 0$. Sum of roots is $\mathbf{2}$.
[Problem 08] Locus of the Center of an Axis-Tangent Circle
Find the equation of the locus of the center $P(x, y)$ of a circle that passes through $(0, 2)$ and is tangent to the $x$-axis.
π Solution & Step-by-Step Breakdown
Answer: $x^2 = 4(y - 1)$
[Solution]
1) Tangent to $x$-axis in upper half-plane $\implies r = y$.
2) Equating center distance from $(0, 2)$ to radius:
$$\sqrt{x^2 + (y - 2)^2} = y \implies x^2 + y^2 - 4y + 4 = y^2 \implies x^2 = 4(y - 1)$$
Equal distance from a fixed point and a fixed line is the geometric definition of a parabola. Squaring both sides yields the locus in two lines.
π― [Self-Diagnosis Variant 08] (Locus along the $y$-Axis)
Find the locus equation of the center of a circle passing through $(4, 0)$ and tangent to the $y$-axis.
View Variant Solution
Breakdown: $(x - 4)^2 + y^2 = x^2 \implies \mathbf{y^2 = 8(x - 2)}$.
[Problem 09] Mutually Externally Tangent Circles on a Shared Axis
Two circles $C_1: (x - 2)^2 + (y - 3)^2 = 9$ and $C_2: (x - 8)^2 + (y - r)^2 = r^2$ are both tangent to the $x$-axis and touch each other externally. Find $r$.
π Solution & Step-by-Step Breakdown
Answer: $3$
[Solution]
1) Center distance equals sum of radii: $d = r_1 + r_2 \implies \sqrt{6^2 + (r - 3)^2} = r + 3$.
2) Squaring: $36 + (r - 3)^2 = (r + 3)^2 \implies 12r = 36 \implies r = \mathbf{3}$.
For two circles touching a common tangent line externally, the horizontal distance between centers is always $\Delta x = 2\sqrt{r_1 r_2}$. Here, $6 = 2\sqrt{3r} \implies r = 3$.
π― [Self-Diagnosis Variant 09] (External Tangency Horizontal Separation)
A circle centered at $(1, 2)$ touches the $x$-axis and externally touches another $x$-axis tangent circle centered at $(7, r)$. Find $r$.
View Variant Solution
Breakdown: $\Delta x = 6 = 2\sqrt{2r} \implies \sqrt{2r} = 3 \implies 2r = 9 \implies r = \mathbf{\dfrac{9}{2}}$.
Theme 3. Simultaneous Dual-Axis Tangency (10 ~ 15)
▲ The centers of two circles passing through a given point $P(x_0, y_0)$ and tangent to both axes lie on $y = x$, with point $P$ being their mutual perimeter intersection.
[Problem 10] Quadrant II Dual-Axis Tangency Distance
Find the distance between the centers of the two circles that pass through $(-2, 1)$ and are tangent to both coordinate axes.
π Solution & Step-by-Step Breakdown
Answer: $4\sqrt{2}$
[Solution]
1) Point $(-2, 1)$ is in Quadrant II $\implies$ Center $(-r, r)$ with $(x + r)^2 + (y - r)^2 = r^2$.
2) Substitute $(-2, 1)$:
$$(-2 + r)^2 + (1 - r)^2 = r^2 \implies r^2 - 6r + 5 = 0 \implies r = 1, 5$$
Centers are $C_1(-1, 1)$ and $C_2(-5, 5)$.
Distance $= \sqrt{(-4)^2 + 4^2} = \mathbf{4\sqrt{2}}$.
Always verify quadrant coordinates first. Because centers lie along diagonal lines $y = \pm x$, the distance between dual centers is simply $\sqrt{2}|r_2 - r_1|$.
π― [Self-Diagnosis Variant 10] (Quadrant IV Dual-Axis Tangency)
Find the distance between the centers of two circles passing through $(2, -1)$ and tangent to both axes.
View Variant Solution
Breakdown: Center $(r, -r)$. $(2 - r)^2 + (-1 + r)^2 = r^2 \implies r = 1, 5$. Distance: $\sqrt{2}(5 - 1) = \mathbf{4\sqrt{2}}$.
[Problem 11] Total Circle Count with Centers along a Parabola
Find the total number of circles whose centers lie on $y = x^2 - 2$ and are tangent to both the $x$-axis and $y$-axis.
π Solution & Step-by-Step Breakdown
Answer: $4$
[Solution]
Centers must lie on either $y = x$ or $y = -x$.
- $x^2 - 2 = x \implies x^2 - x - 2 = 0 \implies x = 2, -1$ ($2$ circles).
- $x^2 - 2 = -x \implies x^2 + x - 2 = 0 \implies x = -2, 1$ ($2$ circles).
Total circles $= 2 + 2 = \mathbf{4}$.
Do not construct circle equations. Count real intersections between $y = f(x)$ and $y = \pm x$.
π― [Self-Diagnosis Variant 11] (Intersection Count via Discriminant)
Find $k$ such that exactly $3$ circles tangent to both axes have their centers on $y = x^2 + k$.
View Variant Solution
Breakdown: Tangency with $y = x \implies D = 1 - 4k = 0 \implies k = \mathbf{\dfrac{1}{4}}$.
[Problem 12] Common Chord of Two Circles Tangent to Both Axes
Find the length of the common chord of the two circles passing through $(1, 2)$ that touch both axes in Quadrant I.
π Solution & Step-by-Step Breakdown
Answer: $\sqrt{2}$
[Solution]
1) $r = 1, 5 \implies C_1: x^2 + y^2 - 2x - 2y + 1 = 0, C_2: x^2 + y^2 - 10x - 10y + 25 = 0$.
2) Common chord ($C_2 - C_1 = 0$): $x + y - 3 = 0$.
3) Distance from $(1, 1)$ to chord: $d = \dfrac{1}{\sqrt{2}}$.
4) Length $= 2\sqrt{r^2 - d^2} = 2\sqrt{1 - 1/2} = \mathbf{\sqrt{2}}$.
The common chord of two circles with centers on $y = x$ is perpendicular to $y = x$, meaning its slope is always $-1$. Subtracting general forms gives the chord equation immediately.
π― [Self-Diagnosis Variant 12] (Common Chord Equation)
Find the equation of the common chord of the two circles passing through $(2, 4)$ and touching both axes in Quadrant I.
View Variant Solution
Breakdown: $r = 2, 10$. Subtracting general forms: $-16x - 16y + 96 = 0 \implies \mathbf{x + y - 6 = 0}$.
[Problem 13] Exterior Tangency for Dual-Axis Circles
Find the radius of the larger circle that touches both axes and is externally tangent to a circle centered at $(3, 3)$ that also touches both axes.
π Solution & Step-by-Step Breakdown
Answer: $3(3 + 2\sqrt{2})$
[Solution]
Center distance $d = R + r \implies \sqrt{2}(R - 3) = R + 3 \implies (\sqrt{2} - 1)R = 3(\sqrt{2} + 1)$.
$R = 3(\sqrt{2} + 1)^2 = \mathbf{3(3 + 2\sqrt{2})}$.
When both circles lie along $y = x$, the center distance is $\sqrt{2}(R - r)$. Setting this equal to the external tangency condition $R + r$ isolates $R$ in one step.
π― [Self-Diagnosis Variant 13] (Tangent Circles with Fixed Center Scale)
Find the radius of a circle tangent to both axes that externally touches a circle centered at $(1, 1)$ with $r = 1$.
View Variant Solution
Breakdown: $(\sqrt{2}-1)R = \sqrt{2}+1 \implies R = (\sqrt{2}+1)^2 = \mathbf{3 + 2\sqrt{2}}$.
[Problem 14] Area Ratio via Similarity Transformations
Find the ratio of the areas of the two circles that pass through $(3, 6)$ and touch both axes in Quadrant I.
π Solution & Step-by-Step Breakdown
Answer: $1 : 25$
[Solution]
1) $(3 - r)^2 + (6 - r)^2 = r^2 \implies r^2 - 18r + 45 = 0 \implies r = 3, 15$.
2) The linear scale ratio is $3 : 15 = 1 : 5$.
Area ratio $= 1^2 : 5^2 = \mathbf{1 : 25}$.
All circles are similar geometric figures. Area ratios equal the square of the ratio of their radii: $(\text{radius ratio})^2$.
π― [Self-Diagnosis Variant 14] (Radii Ratio from Given Point)
Find the ratio of the radii of the two circles passing through $(1, 3)$ that touch both coordinate axes.
View Variant Solution
Breakdown: $r^2 - 8r + 10 = 0 \implies r = 4 \pm \sqrt{6}$. Ratio: $\mathbf{(4 - \sqrt{6}) : (4 + \sqrt{6})}$.
[Problem 15] Single Intersection Condition with a Line
Find the positive constant $k$ such that the line $3x - 4y + k = 0$ intersects the circle centered at $(2, 2)$ (which touches both axes) at exactly one point.
π Solution & Step-by-Step Breakdown
Answer: $12$
[Solution]
1) Center $(2, 2)$, radius $r = 2$.
2) Exactly one intersection point means tangency: distance $d = r$.
$$d = \dfrac{|3(2) - 4(2) + k|}{\sqrt{3^2 + (-4)^2}} = \dfrac{|k - 2|}{5} = 2 \implies |k - 2| = 10$$
Since $k > 0$, $k - 2 = 10 \implies k = \mathbf{12}$.
Do not substitute $y = \dots$ into the circle equation to set discriminant $D = 0$. The geometric identity $d = r$ reduces algebraic workload by over 80%.
π― [Self-Diagnosis Variant 15] (Distance $d = r$ Tangency Setup)
Find $k$ such that $4x + 3y + k = 0$ touches the circle centered at $(1, 1)$ (tangent to both axes) at a single point.
View Variant Solution
Breakdown: $d = \dfrac{|7 + k|}{5} = 1 \implies |7 + k| = 5 \implies k = \mathbf{-2 \text{ or } -12}$.
Students rarely struggle with circles because the formulas are hard; they struggle with misplaced quadrant signs or exhaustion from tedious fractional square completions.
Remember: "The origin circle $x^2 + y^2 = r^2$ is the prototype, and axis tangency simply rolls that prototype until it rests against the walls of the plane."
Once these 15 geometric principles become second nature, the upcoming masterclass—[Circle Concept ②: Skip the Discriminant, $d = r$ & The 3 Tangent Line Formulas]—will feel effortless and clear.
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