[Transformations: Practice Lab ②] Reflections, Line Symmetry & Point Symmetry: 15 Advanced Killer Problems & Self-Diagnosis
Following our study in [Transformations: Masterclass ②], where we explored axial reflections across $x$-axis, $y$-axis, the origin, lines $y = \pm x$, line reflections across $x = a, y = b$, and the 3-second point symmetry formula $(2a - x, 2b - y)$, we now advance to the real arena of competitive and standardized problem solving.
High-level examinations (such as AP Precalculus, SAT Math Module 2, IB Math AA HL, and AMC competitions) do not test trivial sign substitutions. Instead, they require you to dismantle composite mappings $f(-y + 1, x - 2) = 0$ in reverse order, calculate reflections across general linear axes $ax + by + c = 0$ via perpendicular and midpoint constraints, locate geometric axes of symmetry for conic sections, and master Heron's shortest reflection paths.
In this Practice Lab, we break down 15 Advanced Killer Problems alongside their 1:1 Parallel Self-Diagnosis Variants. Sketch each axis on scratch paper, test your algebraic steps, and open the toggles to examine the solutions.
1. Composite Forms $f(-y + a, x + b) = 0$ Demand 'Reflect First, Translate Second': Execute coordinate swaps ($y = x$ reflection) and sign inversions (axial reflections) first, then factor out leading coefficients to read pure translation displacements.
2. The 2 Invariant Conditions for Oblique Line Reflections: When reflecting across an arbitrary line $l$, enforce only two conditions: [Perpendicularity: slope product $= -1$] and [Midpoint Constraint: midpoint of segment lies on $l$].
3. Unfold Piecewise Shortest Paths Across 'Mirror Boundaries': Treat the line containing moving point $P$ as an optical mirror. Reflecting a fixed point across this boundary straightens bent paths into a single Euclidean segment.
Theme 1. Fundamental Reflections & Composite Mappings (01 ~ 04)
[Problem 01] Deconstructing Composite Transformations in Sequential Order
The curve $f(x, y) = 0$ is reflected across the line $y = x$, then translated horizontally by $+2$ and vertically by $-1$, and finally reflected across the $x$-axis to yield equation $g(x, y) = 0$. Express $g(x, y) = 0$ in terms of $f$.
π Solution & Step-by-Step Breakdown
Answer: $f(-y + 1, x - 2) = 0$
[Step-by-Step Substitution Trace]
Step 1 (Reflection across $y = x$): Swap the names of variables $x$ and $y$:
$$f(x, y) = 0 \longrightarrow f(y, x) = 0$$
Step 2 (Translation by $+2$ along $x$-axis and $-1$ along $y$-axis): Substitute $(x - 2)$ for the currently appearing variable $x$, and $(y + 1)$ for the currently appearing variable $y$:
$$f(y, x) = 0 \longrightarrow f(y + 1, x - 2) = 0$$
Step 3 (Reflection across the $x$-axis): Replace the currently appearing variable $y$ with $-y$:
$$f(y + 1, x - 2) = 0 \longrightarrow f(-y + 1, x - 2) = 0$$
Thus, the final governing equation is $\mathbf{f(-y + 1, x - 2) = 0}$.
Do not think in terms of "first slot / second slot." Target the physical variable symbols $x$ and $y$ as they currently appear in the expression and apply substitutions directly.
π― [Self-Diagnosis Variant 01] (Composite Inversion Sequence)
The curve $f(x, y) = 0$ is reflected across the $y$-axis, translated horizontally by $-3$, and then reflected across $y = -x$. Find the resulting equation.
View Variant Solution
Breakdown: $f(x, y)=0 \xrightarrow{y\text{-axis}} f(-x, y)=0 \xrightarrow{\Delta x=-3} f(-(x+3), y)=f(-x-3, y)=0 \xrightarrow{y=-x} f(-(-y)-3, -x) = \mathbf{f(y - 3, -x) = 0}$.
[Problem 02] Common Tangent to a Parabola and its Origin Reflection
Parabola $C_1: y = x^2 - 4x + 5$ is reflected across the origin to form parabola $C_2$. Determine the equation of the common tangent line with positive slope that touches both $C_1$ and $C_2$.
π Solution & Step-by-Step Breakdown
Answer: $y = (2\sqrt{5} - 4)x$
[Solution]
1) $C_1$ standard form: $y = (x - 2)^2 + 1 \implies \text{Vertex: } (2, 1)$, opens upward ($a = 1$).
2) Reflecting across the origin gives $C_2$: vertex $(-2, -1)$, opens downward ($a = -1$):
$$-y = (-x)^2 - 4(-x) + 5 \implies y = -x^2 - 4x - 5$$
3) Let the common tangent be $y = mx + n$ ($m > 0$). Enforce tangency via discriminant $D = 0$ on both curves:
• $x^2 - (4 + m)x + (5 - n) = 0 \implies D_1 = (m + 4)^2 - 4(5 - n) = 0$
• $x^2 + (m + 4)x + (n + 5) = 0 \implies D_2 = (m + 4)^2 - 4(n + 5) = 0$
4) Equating discriminants: $5 - n = n + 5 \implies n = 0$ (the tangent line passes directly through the origin).
5) Substituting $n = 0$ into $D_1$: $(m + 4)^2 = 20 \implies m + 4 = 2\sqrt{5} \implies m = 2\sqrt{5} - 4$.
Thus, the common tangent is $\mathbf{y = (2\sqrt{5} - 4)x}$.
Any non-vertical line that is simultaneously tangent to two point-symmetric curves about the origin must pass through the origin itself ($n = 0$). Recognizing this symmetry cuts algebra by two-thirds.
π― [Self-Diagnosis Variant 02] (Intersection of Curve and its Inverse)
Determine the intersection points between $y = x^2 + 1$ and its reflection across $y = x$.
View Variant Solution
Breakdown: Intersecting $y = x^2 + 1$ with line of symmetry $y = x$ yields $x^2 - x + 1 = 0$, which has no real roots ($D < 0$).
[Problem 03] Reflection of a Circle Across $y = -x$ Tangent to Both Axes
Circle $C: (x - a)^2 + (y - 2)^2 = 4$ is reflected across the line $y = -x$ to form $C'$. Find the positive constant $a$ such that $C'$ is simultaneously tangent to both coordinate axes.
π Solution & Step-by-Step Breakdown
Answer: $a = 2$
[Solution]
1) Center of $C$ is $(a, 2)$, radius $r = 2$.
2) Reflection across $y = -x$ maps $(x, y) \to (-y, -x)$.
The new center $O'$ is $(-2, -a)$, while radius remains $r = 2$.
3) For a circle to touch both axes simultaneously, both center coordinates must have an absolute value equal to radius $r = 2$:
$$|-2| = 2 \quad \text{and} \quad |-a| = 2 \implies \mathbf{a = 2}$$
Remember that $y = -x$ swaps coordinates AND negates both signs: $(x, y) \to (-y, -x)$. For axis tangency, enforce $|\text{Center } x| = |\text{Center } y| = r$.
π― [Self-Diagnosis Variant 03] (Common Chord of Reflected Circles)
Find the common chord of the circle $(x - 2)^2 + y^2 = 2$ reflected across the $y$-axis and across $y = x$.
View Variant Solution
Breakdown: Subtracting $(x + 2)^2 + y^2 = 2$ and $x^2 + (y - 2)^2 = 2$ gives $4x + 4y = 0 \implies \mathbf{x + y = 0}$.
[Problem 04] Area Enclosed by Repeated Four-Way Reflections of a Line Segment
Line segment $AB$ connecting $A(2, 0)$ and $B(0, 3)$ in the first quadrant is reflected across the $x$-axis, the $y$-axis, and the origin. Find the total area enclosed by the resulting closed figure.
π Solution & Step-by-Step Breakdown
Answer: $12$
[Solution]
1) Segment $AB$ represents the first-quadrant portion of line $\dfrac{x}{2} + \dfrac{y}{3} = 1$.
2) Applying reflections across both axes generates the rhombus $\dfrac{|x|}{2} + \dfrac{|y|}{3} = 1$.
3) Area in quadrant 1 is $S_1 = \dfrac{1}{2} \times 2 \times 3 = 3$.
4) By fourfold symmetry, total enclosed area is $4 \times 3 = \mathbf{12}$.
Simultaneous $x$-axis and $y$-axis reflection wraps variables in absolute values: $|x|, |y|$. Compute the single triangle in Quadrant 1 and multiply by 4.
π― [Self-Diagnosis Variant 04] (Symmetric Absolute Diamond Area)
Find the area enclosed by $y = -|x| + 4$ ($y \ge 0$) and its reflection across the $x$-axis.
View Variant Solution
Breakdown: Rhombus with diagonal lengths $d_1 = 8, d_2 = 8$. Area $= \dfrac{1}{2} \times 8 \times 8 = \mathbf{32}$.
Theme 2. Symmetries Across Lines $x = a$ and $y = b$ (05 ~ 07)
[Problem 05] External Tangency Locus Under Vertical Line Symmetry
Circle $C_1: (x + 2)^2 + (y - 3)^2 = 4$ is reflected across the line $x = 1$ to form circle $C_2$. Determine the $x$-coordinate of the center of a circle of radius $1$ that is externally tangent to both $C_1$ and $C_2$.
π Solution & Step-by-Step Breakdown
Answer: $x = 1$
[Solution]
1) Center of $C_1$ is $O_1(-2, 3)$, radius $r_1 = 2$.
2) Reflecting $O_1$ across line $x = 1$ gives center $O_2$ of circle $C_2$:
$$x' = 2(1) - (-2) = 4, \quad y' = 3 \implies O_2(4, 3)$$
3) The perpendicular bisector of segment $\overline{O_1 O_2}$ is the line of symmetry $x = 1$.
4) Any circle tangent to two congruent circles must have its center equidistant from both centers, which places it on the perpendicular bisector $x = 1$.
Therefore, the $x$-coordinate of the center is strictly $\mathbf{x = 1}$.
When two shapes are symmetric across $x = a$, any mutually tangent circle must center along the axis of symmetry. Eliminate unknown variables immediately by symmetry.
π― [Self-Diagnosis Variant 05] (Horizontal Line Reflection Intersections)
Find the sum of the $x$-coordinates of the intersections between $y = (x - 1)^2 + 2$ and its reflection across $y = 5$.
View Variant Solution
Breakdown: Both curves share vertical axis $x = 1$. The sum of symmetric roots is $2 \times 1 = \mathbf{2}$.
[Problem 06] Symmetry Identity $f(2 - x) = f(4 + x)$ in Quadratic Extrema
Quadratic function $f(x) = x^2 - ax + b$ satisfies $f(2 - x) = f(4 + x)$ for all $x \in \mathbb{R}$, and its minimum value is $-3$. Find $a + b$.
π Solution & Step-by-Step Breakdown
Answer: $12$
[Solution]
1) Add the inputs: $(2 - x) + (4 + x) = 6$ (constant).
The axis of symmetry is the midpoint: $x = \dfrac{6}{2} = 3$.
2) Standard form for quadratic opening upward with vertex at $(3, -3)$:
$$f(x) = (x - 3)^2 - 3 = x^2 - 6x + 6$$
3) Equating coefficients: $a = 6, b = 6 \implies a + b = \mathbf{12}$.
Whenever you see $f(p - x) = f(q + x)$, sum the arguments. The variable $x$ cancels out, and the axis of symmetry is always $x = \dfrac{p + q}{2}$.
π― [Self-Diagnosis Variant 06] (Cubic Extrema Under Line Symmetry)
Function $y = f(x)$ satisfies $f(1 - x) = f(5 + x)$ and possesses two local extrema. Find the sum of the $x$-coordinates of these extrema.
View Variant Solution
Breakdown: Axis of symmetry is $x = \dfrac{1 + 5}{2} = 3$. Sum of symmetric roots is $2 \times 3 = \mathbf{6}$.
[Problem 07] Composition of Two Parallel Line Reflections
Curve $f(x, y) = 0$ is reflected across the line $x = 1$, and then reflected across the line $x = 4$. Determine the equivalent single pure translation.
π Solution & Step-by-Step Breakdown
Answer: Translation of $+6$ along the $x$-axis
[Solution]
1) First reflection across $x = 1$: $x \to 2(1) - x = 2 - x \implies f(2 - x, y) = 0$.
2) Second reflection across $x = 4$: substitute $2(4) - x = 8 - x$ into variable $x$:
$$f(2 - (8 - x), y) = 0 \implies f(x - 6, y) = 0$$
3) Replacing $x$ with $x - 6$ corresponds to a pure horizontal translation of $+6$.
Notice this equals twice the distance between the lines: $2 \times (4 - 1) = 6$.
Reflecting across two parallel lines separated by distance $d$ produces a pure translation of $2d$ in the direction perpendicular to the mirrors.
π― [Self-Diagnosis Variant 07] (Consecutive Horizontal Line Reflections)
Curve $f(x, y) = 0$ is reflected across $y = -2$ and then across $y = 3$. Find the vertical translation amount.
View Variant Solution
Breakdown: Axis separation is $3 - (-2) = 5$. Translation amount is $2 \times 5 = \mathbf{+10}$.
Theme 3. Point Symmetry $P(a, b)$ & Functional Equations (08 ~ 10)
[Problem 08] Symmetric Pairing in $f(1 + x) + f(3 - x) = 6$
Function $f(x)$ satisfies $f(1 + x) + f(3 - x) = 6$ for all $x \in \mathbb{R}$. Evaluate $\sum_{k=1}^{3} f(k) = f(1) + f(2) + f(3)$.
π Solution & Step-by-Step Breakdown
Answer: $9$
[Solution]
1) The sum of arguments is $(1 + x) + (3 - x) = 4$, indicating point symmetry about center $(2, 3)$.
2) Setting $x = 0$ gives $f(1) + f(3) = 6$.
3) Setting $x = 1$ gives $f(2) + f(2) = 6 \implies f(2) = 3$.
4) Summing: $\{f(1) + f(3)\} + f(2) = 6 + 3 = \mathbf{9}$.
Any function symmetric about $(a, b)$ satisfies $f(a - k) + f(a + k) = 2b$, with $f(a) = b$. Pair terms symmetrically around the center to compute finite sums instantly.
π― [Self-Diagnosis Variant 08] (Point-Symmetric Sequence Sum)
Function $f(x)$ satisfies $f(x) + f(10 - x) = 8$. Evaluate $\sum_{k=1}^{9} f(k)$.
View Variant Solution
Breakdown: Symmetric about $(5, 4)$. Four symmetric pairs evaluate to $4 \times 8 = 32$, plus center $f(5) = 4 \implies \mathbf{36}$.
[Problem 09] Center of Symmetry Between Two Circles
Circle $C_1: x^2 + y^2 - 4x + 6y + 4 = 0$ and circle $C_2: x^2 + y^2 + 8x - 2y + k = 0$ are symmetric with respect to point $P$. Find the coordinates of $P$ and the constant $k$.
π Solution & Step-by-Step Breakdown
Answer: $P(-1, -1), k = 8$
[Solution]
1) Extract centers and radii:
$C_1: (x - 2)^2 + (y + 3)^2 = 9 \implies O_1(2, -3), r_1^2 = 9$
$C_2: (x + 4)^2 + (y - 1)^2 = 17 - k \implies O_2(-4, 1), r_2^2 = 17 - k$
2) Point symmetry preserves radius: $9 = 17 - k \implies \mathbf{k = 8}$.
3) Point $P$ is the midpoint of centers $O_1$ and $O_2$:
$$P = \left(\dfrac{2 - 4}{2}, \dfrac{-3 + 1}{2}\right) = \mathbf{(-1, -1)}$$
When two shapes are point-symmetric, the center of symmetry is the midpoint of their characteristic anchor points, and all shape metrics (radius, side lengths) remain invariant.
π― [Self-Diagnosis Variant 09] (Parabola Vertices Point-Symmetry Center)
Parabolas $y = x^2 - 2x + 3$ and $y = -x^2 + 6x - 1$ are symmetric about point $P(a, b)$. Find $a + b$.
View Variant Solution
Breakdown: Midpoint of vertices $(1, 2)$ and $(3, 8)$ is $P(2, 5) \implies 2 + 5 = \mathbf{7}$.
[Problem 10] Distance Between Parallel Lines Under Point Reflection
Line $l: 2x - y + 4 = 0$ is reflected across point $P(2, 1)$ to form line $l'$. Find the distance between $l$ and $l'$.
π Solution & Step-by-Step Breakdown
Answer: $\dfrac{14\sqrt{5}}{5}$
[Solution]
1) Reflecting a line across a point preserves slope, yielding an identical parallel line.
2) Substitute $x \to 4 - x$ and $y \to 2 - y$:
$$2(4 - x) - (2 - y) + 4 = 0 \implies 2x - y - 10 = 0$$
3) Distance between parallel lines $2x - y + 4 = 0$ and $2x - y - 10 = 0$:
$$d = \dfrac{|4 - (-10)|}{\sqrt{2^2 + (-1)^2}} = \dfrac{14}{\sqrt{5}} = \mathbf{\dfrac{14\sqrt{5}}{5}}$$
The distance between a line and its reflection across point $P$ is twice the distance from $P$ to the line: $2 \times d(P, l)$.
π― [Self-Diagnosis Variant 10] (Origin-Symmetric Parallel Line Distance)
Find the distance between line $3x + 4y - 15 = 0$ and its reflection across the origin.
View Variant Solution
Breakdown: Origin distance is $\dfrac{|-15|}{5} = 3$. Doubling yields $2 \times 3 = \mathbf{6}$.
Theme 4. Oblique Line Reflections Across $ax + by + c = 0$ (11 ~ 13)
[Problem 11] Point Reflection Across a General Line $2x - y + 1 = 0$
Find the coordinates of image $A'(p, q)$ when point $A(1, 4)$ is reflected across line $l: 2x - y + 1 = 0$.
π Solution & Step-by-Step Breakdown
Answer: $\left(\dfrac{9}{5}, \dfrac{18}{5}\right)$
[Solution]
1) [Perpendicular Condition]: Slope of $\overline{AA'}$ is perpendicular to line slope ($2$):
$$\dfrac{q - 4}{p - 1} = -\dfrac{1}{2} \implies p + 2q = 9 \quad \cdots \text{①}$$
2) [Midpoint Condition]: Midpoint $\left(\dfrac{p + 1}{2}, \dfrac{q + 4}{2}\right)$ lies on $2x - y + 1 = 0$:
$$2\left(\dfrac{p + 1}{2}\right) - \left(\dfrac{q + 4}{2}\right) + 1 = 0 \implies 2p - q = 0 \implies q = 2p \quad \cdots \text{②}$$
3) Solving system: $p + 2(2p) = 9 \implies p = \dfrac{9}{5}, q = \dfrac{18}{5}$.
Thus, $A' = \mathbf{\left(\dfrac{9}{5}, \dfrac{18}{5}\right)}$.
For lines with slopes other than $\pm 1$, avoid memorizing shortcut formulas. Solve the 2-by-2 linear system from perpendicularity ($m_1 m_2 = -1$) and the midpoint constraint.
π― [Self-Diagnosis Variant 11] (Circle Center Reflected Across Oblique Line)
Find the center of circle $(x - 1)^2 + (y - 2)^2 = 5$ reflected across line $x + 2y - 10 = 0$.
View Variant Solution
Breakdown: Setting up perpendicular and midpoint conditions for center $(1, 2)$ yields image $\mathbf{(3, 6)}$.
[Problem 12] Reflecting a Line Across Another Line via Invariant Points
Find the equation of line $l_2$ obtained by reflecting line $l_1: x - y + 1 = 0$ across line $m: x - 2y + 3 = 0$.
π Solution & Step-by-Step Breakdown
Answer: $x - 7y + 9 = 0$
[Solution - Two-Point Tracking Method]
1) Find the invariant intersection $P$ between $l_1$ and $m$:
$$x - y + 1 = 0 \quad \text{and} \quad x - 2y + 3 = 0 \implies P(1, 2)$$
Point $P(1, 2)$ lies on mirror line $m$, so it remains fixed under reflection.
2) Pick a simple second point on $l_1$, say $Q(-1, 0)$, and reflect it across $m$ to find $Q'\left(\dfrac{3}{5}, \dfrac{16}{5}\right)$.
3) Connect $P(1, 2)$ and $Q'$ to establish the reflected line equation: $\mathbf{x - 7y + 9 = 0}$.
Do not derive locus equations for line-across-line reflections. Find the fixed intersection point $P$ first, reflect one convenient secondary point $Q$, and join them.
π― [Self-Diagnosis Variant 12] (Angle Bisector Slope Inversion)
Find the slope of line $y = 2x$ reflected across $y = x$.
View Variant Solution
Breakdown: Swapping variables gives $x = 2y \implies y = \dfrac{1}{2}x$. Slope is the reciprocal $\mathbf{\dfrac{1}{2}}$.
[Problem 13] Parabola Vertex Tracking Across Lines of Slope $\pm 1$
Find the vertex coordinates of parabola $y = x^2$ reflected across the line $x - y + 2 = 0$.
π Solution & Step-by-Step Breakdown
Answer: $(-2, 2)$
[Solution]
1) The original parabola has vertex $O(0, 0)$.
2) The mirror line $y = x + 2$ has slope $1$. For lines with slope $\pm 1$, solve directly for $x$ and $y$:
$$x = y - 2, \quad y = x + 2$$
3) Substitute the vertex coordinates $(0, 0)$:
$$x' = 0 - 2 = -2, \quad y' = 0 + 2 = 2$$
The reflected vertex is located at $\mathbf{(-2, 2)}$.
Exclusively when the mirror line has slope $m = \pm 1$, solve the line's equation for $x$ and $y$ to get your transformation formulas directly without solving simultaneous equations.
π― [Self-Diagnosis Variant 13] (Direct Substitution Across $x + y = 3$)
Find the reflection of point $P(4, 1)$ across line $x + y = 3$.
View Variant Solution
Breakdown: $x' = 3 - 1 = 2$, $y' = 3 - 4 = -1 \implies \mathbf{(2, -1)}$.
Theme 5. Heron's Shortest Paths & Folded Circles (14 ~ 15)
[Problem 14] Minimum Triangle Perimeter with Two Moving Vertices (Heron's Path)
Point $A(2, 5)$ is given alongside point $P$ moving along the $x$-axis and point $Q$ moving along the line $y = x$. Find the minimum perimeter of triangle $APQ$: $\overline{AP} + \overline{PQ} + \overline{QA}$.
π Solution & Step-by-Step Breakdown
Answer: $\sqrt{58}$
[Solution - Two-Sided Mirror Reflection]
1) Moving point $P$ lies on the $x$-axis. Reflect $A(2, 5)$ across the $x$-axis to get $A_1(2, -5)$, ensuring $\overline{AP} = \overline{A_1 P}$.
2) Moving point $Q$ lies on $y = x$. Reflect $A(2, 5)$ across line $y = x$ to get $A_2(5, 2)$, ensuring $\overline{QA} = \overline{QA_2}$.
3) Unfold the perimeter into a straight path between the two virtual images:
$$\overline{AP} + \overline{PQ} + \overline{QA} = \overline{A_1 P} + \overline{PQ} + \overline{QA_2} \ge \overline{A_1 A_2}$$
4) The minimum is the Euclidean distance between $A_1(2, -5)$ and $A_2(5, 2)$:
$$\overline{A_1 A_2} = \sqrt{(5 - 2)^2 + (2 - (-5))^2} = \sqrt{3^2 + 7^2} = \mathbf{\sqrt{58}}$$
When two vertices are free to move on distinct boundary lines, reflect the single fixed vertex across both boundary mirrors independently, then measure the straight-line distance between the two reflections.
π― [Self-Diagnosis Variant 14] (Shortest Distance to a Circle via Reflection)
Find the minimum value of $\overline{AQ} + \overline{QP}$ where $A(0, 2)$, point $Q$ moves on the $x$-axis, and $P$ moves on circle $(x - 6)^2 + (y - 8)^2 = 4$.
View Variant Solution
Breakdown: Reflect $A(0, 2)$ across the $x$-axis to $A'(0, -2)$. Distance to circle center $(6, 8)$ is $d = \sqrt{6^2 + 10^2} = 2\sqrt{34}$. Minimum distance is $d - r = \mathbf{2\sqrt{34} - 2}$.
[Problem 15] Reconstructing the Crease Line of a Folded Circular Arc
Circle $C: x^2 + y^2 = 25$ is folded along crease line $l$ such that the folded circular arc is tangent to the $x$-axis at point $P(1, 0)$. Find the equation of the reflected circle and the equation of crease line $l$.
π Solution & Step-by-Step Breakdown
Answer: Reflected circle: $(x - 1)^2 + (y - 5)^2 = 25$; Crease: $x + 5y - 13 = 0$
[Solution - Full Circle Reconstruction Principle]
1) Do not treat the folded arc in isolation; complete it into a congruent virtual circle $C'$ having radius $r = 5$.
2) Since the arc touches the $x$-axis at $(1, 0)$, the new center $O'$ must have $x = 1$ and $y = 5$ (in Quadrant 1):
$$C': \mathbf{(x - 1)^2 + (y - 5)^2 = 25}$$
3) The crease line $l$ is the perpendicular bisector of segment $\overline{OO'}$ connecting original center $O(0, 0)$ to new center $O'(1, 5)$:
Midpoint is $\left(\dfrac{1}{2}, \dfrac{5}{2}\right)$, perpendicular slope is $-\dfrac{1}{5}$.
$$y - \dfrac{5}{2} = -\dfrac{1}{5}\left(x - \dfrac{1}{2}\right) \iff \mathbf{x + 5y - 13 = 0}$$
Any circular fold problem is solved by completing the full circle of identical radius. The fold line is always the perpendicular bisector between the original center and the virtual center.
π― [Self-Diagnosis Variant 15] (Folded Arc Passing Through Fixed Point)
Circle $x^2 + y^2 = 10$ is folded along a chord such that the arc passes through $(0, 2)$. Find the minimum slope of the crease line.
View Variant Solution
Breakdown: Set virtual circle $(x - a)^2 + (y - b)^2 = 10$ passing through $(0, 2)$ to trace center $(a, b)$, then derive the extrema of the bisector slope.
Mastery over reflections is not about memorizing algebraic signs—it is about seeing coordinate space as an unfolded geometric plane.
When you reflect across mirrors to straighten piecewise paths, track virtual centers instead of isolated arcs, and deploy $(2a - x, 2b - y)$ automatically, even the most formidable transformation problems unlock effortlessly.
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