Invariant Power Slides Defining Power of a Point
Metric Relationships in Circle Geometry
Lesson 01 // Undergraduate Geometry
The Geometric Magic Trick
Pick any point P inside a circle. Draw any chord through P.
Label the intersections A and B.
PA ⋅ PB = Constant?
No matter the orientation of the chord, the product of these lengths never changes. Why?
P A B
Formal Definition
Let Γ be a circle with center O and radius r. The Power of a Point P with respect to Γ is defined as:
\[ \mathcal{P}_{\Gamma}(P) = d^2 - r^2 \]
where d = dist(P, O)
Inside
Negative
\( d < r \)
On Circle
Zero
\( d = r \)
Outside
Positive
\( d > r \)
The Power of a Point Theorem
"For any line passing through P that intersects the circle at A and B, the product of signed distances \( \vec{PA} \cdot \vec{PB} \) is invariant."
Independent of the line's orientation.
Connects directly to the quadratic form of the circle.
Sets the stage for radical axis exploration.
Algebraic Link
If Γ is \( x^2 + y^2 = r^2 \)
and \( P = (x_0, y_0) \)
\( \mathcal{P}(P) = x_0^2 + y_0^2 - r^2 \)
Dynamic Exploration Task
Open your dynamic geometry software (GeoGebra/Desmos). Construct a circle and a point P. Measure the segments PA and PB as you rotate the line.
Investigate:
1 What happens to the product when P moves closer to the center?
2 When P is outside, how does this relate to the tangent length?
3 Can you find a point with Power = Radius?
Point Power Teacher Guide Point Power Teacher Guide
Metric Relationships // Lesson 01: Defining Power of a Point
UG-GEO-01
Lesson Context
This introductory lesson shifts the focus from traditional "segment length" theorems to a unified scalar property of a point relative to a circle. By defining the Power of a Point algebraically as \( P = d^2 - r^2 \), students can later derive the intersecting chords, secants, and tangent theorems as special cases of a single invariant.
Key Concepts
Geometric Invariance
Scalar Power
Vector/Signed Products
Facilitation Guide
1
The "Constant Product" Inquiry (15 mins)
Start with a dynamic geometry simulation. Ask students to measure the lengths of segments for chords through a fixed interior point. Students should discover that while the individual lengths change drastically during rotation, the product is constant.
Guiding Question:
"If we move the point P closer to the center, does the product increase or decrease? Why?"
2
The Formal Definition (20 mins)
Introduce the algebraic definition \( \mathcal{P}(P) = d^2 - r^2 \). Explain the significance of the sign:
Negative: Point is internal (chords).
Zero: Point is on the circumference.
Positive: Point is external (tangents/secants).
3
Connection to Coordinate Geometry (15 mins)
Show that the power of point \( (x_0, y_0) \) with respect to circle \( (x-h)^2 + (y-k)^2 - r^2 = 0 \) is simply the value obtained by substituting the point's coordinates into the circle's equation. This bridge is critical for the upcoming Radical Axis lesson.
Common Misconceptions
Signed Distances: Students often forget that for interior points, \( PA \) and \( PB \) are in opposite directions, making the product \( \vec{PA} \cdot \vec{PB} \) negative.Units: Remind students that Power is an area-like unit (length squared).
Extension Tasks
Challenge advanced students to prove the invariance using the law of cosines for a general line intersecting the circle at an angle \( \theta \).
Metric Relationships Circle Power // Sequence 1, Lesson 1 Guide
Invariant Product Activity The Invariant Product
Undergraduate Geometry // Power of a Point
Name:
Date:
Part 1: The Scalar Power
The Power of a Point \( P \) with respect to a circle \( \Gamma \) of radius \( r \) and center \( O \) is given by \( \mathcal{P}(P) = d^2 - r^2 \), where \( d \) is the distance from \( P \) to \( O \).
1. Calculate the Power of point \( P \) relative to circle \( \Gamma \):
Circle Γ: \( x^2 + y^2 = 25 \)
Point P: \( (3, -2) \)
Show work here
2. Interpretation of the Sign:
Based on your result in Question 1, is the point \( P \) inside, outside, or on the circle? Justify your answer using the sign of the power.
Part 2: The Invariant Product
The Power of a Point theorem states that for any line through \( P \) intersecting the circle at \( A \) and \( B \), the product \( PA \cdot PB \) (signed) is constant.
P A B
3. The Horizontal Diameter Case:
Let point \( P \) lie on the x-axis at distance \( d \) from the center of a circle with radius \( r \). Consider the horizontal line (the x-axis) through \( P \). Find the coordinates of intersections \( A \) and \( B \), and show that \( PA \cdot PB = d^2 - r^2 \).
4. Inquiry Prompt:
If a point \( P \) is outside the circle, we can draw a tangent from \( P \) to the circle at point \( T \). How does the length of the segment \( PT \) relate to the power \( \mathcal{P}(P) \)? Explain your reasoning.
Challenge
Prove that the Power of a Point \( P(x_0, y_0) \) with respect to circle \( x^2 + y^2 + 2gx + 2fy + c = 0 \) is exactly \( x_0^2 + y_0^2 + 2gx_0 + 2fy_0 + c \). Hint: Complete the square to find the radius and center.
Unified Circle Metrics Slides Unifying Segments
Similarity, Symmetry, and the Power Invariant
Lesson 02 // Undergrad Circle Power
The Fragmented View
1. Intersecting Chords
Interior point: \( PA \cdot PB = PC \cdot PD \)
2. Intersecting Secants
Exterior point: \( PA \cdot PB = PC \cdot PD \)
3. Tangent-Secant
One tangent: \( PT^2 = PA \cdot PB \)
"What if these are all the same theorem?"
The Similarity Link
For any two chords intersecting at P, triangles PAC and PDB are similar.
Vertical angles at P are equal.
Angles inscribed in the same arc are equal.
\[ \frac{PA}{PD} = \frac{PC}{PB} \implies PA \cdot PB = PC \cdot PD \]
P A B C D
The Continuity of Form
Continuity Argument
As secant line \( PAB \) rotates, the points \( A \) and \( B \) approach each other.
At the limit, \( A = B = T \), and the product becomes:
\( PT \cdot PT = PT^2 \)
"The tangent is just a secant with coincident intersections."
Unified Result
For any point P, the product \( PA \cdot PB \) is constant and equal to \( d^2 - r^2 \).
Vector Interpretation
Inside the circle, A and B are on opposite sides of P (\( \vec{PA} \cdot \vec{PB} < 0 \)). Outside, they are on the same side (\( \vec{PA} \cdot \vec{PB} > 0 \)).
Challenge Problem
The Power of Symmetry
A point P is 13 units from the center of a circle with radius 5. A secant through P intersects the circle such that the internal segment AB has length 8.
Find the lengths of PA and PB.
Hint 1: Calculate the Power of P first.
Hint 2: Let PA = x and PB = x + 8.
Metric Bridges Teacher Guide Metric Bridges Guide
Metric Relationships // Lesson 02: Unifying Segments
UG-GEO-02
Lesson Purpose
This lesson aims to dismantle the student's perception of circle theorems as three separate rules (Chords, Secants, Tangents). By using similar triangles and continuity arguments, students should arrive at the understanding that all these theorems are manifestations of the Power of a Point invariant.
Proof Focus
Triangle Similarity in Circle Geometry
Limit Cases of Secant Rotation
Vector Continuity (Inside vs Outside)
Proof Facilitation
1
The Chord Proof (Interior)
Remind students of the inscribed angle theorem: angles subtended by the same arc are equal.
Key Step: In triangles PAC and PDB, ∠PAC = ∠PDB (same arc) and ∠APC = ∠DPB (vertical). Thus, ΔPAC ∼ ΔPDB.
2
The Secant-Tangent Continuity
This is the crucial conceptual leap. As a secant through P rotates towards tangency, intersections A and B converge to a single point T.
Teaching Note:
"Stress that the tangent isn't a 'new' geometry, but the limit of the secant geometry. This aligns with calculus concepts (derivatives as limits of secants)."
3
The Unified Scalar (Vector) Power
Introduce the signed distance: \( \vec{PA} \cdot \vec{PB} \).
Interior: A and B are on opposite sides of P. Product is negative.Exterior: A and B are on the same side. Product is positive.
Workshop Key (Slide 5)
Problem: P is 13 from center, r=5. Chord AB = 8. Find PA, PB.
Calculate Power: \( \mathcal{P}(P) = d^2 - r^2 = 13^2 - 5^2 = 169 - 25 = 144 \).Set Up Equation: \( PA \cdot PB = 144 \).Substitution: Let \( PA = x \). Then \( PB = x + 8 \) (since P is external).Solve Quadratic: \( x(x + 8) = 144 \implies x^2 + 8x - 144 = 0 \).Roots: Using the quadratic formula, \( x = \frac{-8 \pm \sqrt{64 + 576}}{2} = \frac{-8 \pm 8\sqrt{10}}{2} = -4 \pm 4\sqrt{10} \).Final Lengths: Since lengths are positive, \( PA = 4\sqrt{10} - 4 \approx 8.65 \), and \( PB = 4\sqrt{10} + 4 \approx 16.65 \).
Metric Relationships Circle Power // Sequence 1, Lesson 2 Guide
The Similarity Link Worksheet The Similarity Link
Metric Unity in Circle Geometry
Name:
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Proof Challenge: Similarity
Consider a circle with two chords \( AB \) and \( CD \) intersecting at an interior point \( P \).
1. Formulate the proof that \( \Delta PAC \sim \Delta PDB \):
a) Identify the equal angles subtended by arc \( CB \):
b) Identify the vertical angles at \( P \):
c) Write the ratio of corresponding sides and cross-multiply to show \( PA \cdot PB = PC \cdot PD \):
P A B C D
Part 2: Tangent-Secant Identity
2. Continuity Argument:
Explain why, as the chord \( CD \) in the previous example rotates around \( P \) until it becomes a tangent at point \( T \), the product \( PC \cdot PD \) becomes \( PT^2 \). Why does this reinforce the definition of Power of a Point for exterior points?
3. Numerical Application:
A point \( P \) has Power 64 with respect to circle \( \Gamma \). A secant from \( P \) passes through the circle, and the length of the chord segment inside the circle is 12. Let \( PA = x \) be the distance from \( P \) to the nearest intersection. Solve for \( x \).
Part 3: The Geometric Scalar
4. Vector Sign Interpretation:
"If we define the power as the dot product of vectors from P to the two intersections, \( \mathcal{P}(P) = \vec{PA} \cdot \vec{PB} \), how does the orientation of the segments change when P moves from the interior to the exterior?"
Internal Case
Sketch & Explain
External Case
Sketch & Explain
Metric Relationships Circle Power // Sequence 1, Lesson 2 Worksheet
Neutral Ground Slides The Radical Axis
The Locus of Equal Power
Lesson 03 // Undergrad Circle Power
The Balance of Power
Given two circles Γ₁ and Γ₂...
Where is the set of all points P that have the exact same power relative to both?
\[ \mathcal{P}_{\Gamma_1}(P) = \mathcal{P}_{\Gamma_2}(P) \]
Spoiler: It's always a straight line.
Radical Axis Γ₁ Γ₂
Algebraic Logic
Let two circles be defined as:
\( C_1: x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0 \)
\( C_2: x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0 \)
Equating their powers at \( (x, y) \):
\( C_1(x,y) = C_2(x,y) \)
The quadratic terms \( x^2 + y^2 \) cancel out, leaving a linear equation.
Equation of Axis
\[ 2(g_1-g_2)x + 2(f_1-f_2)y + (c_1-c_2) = 0 \]
Geometric Properties
Orthogonality
The radical axis is always perpendicular to the line segment connecting the centers of the two circles.
Intersection Case
If circles intersect, the radical axis is the Common Chord. If they are tangent, it is the Common Tangent.
O₁ O₂
Tangent Equality
For any point P on the radical axis outside the circles, the tangents from P to both circles have the exact same length.
\[ PT_1 = PT_2 = \sqrt{\mathcal{P}(P)} \]
This property allows us to construct the radical axis for non-intersecting circles by finding points of equal tangent lengths.
Locus Logic Teacher Guide Locus Logic Guide
Metric Relationships // Lesson 03: The Radical Axis
UG-GEO-03
Lesson Purpose
This lesson expands the concept of Power of a Point from a single circle to a relationship between two circles. Students will define and construct the Radical Axis as the locus of points where the power relative to two circles is equal.
Analytical Focus
Locus as Algebraic Constraint
Linearization of Quadratic Forms
Construction via Tangent Lengths
Instructional Steps
1
The Cancellation Discovery (20 mins)
Write the general equations of two circles \( C_1 = 0 \) and \( C_2 = 0 \). Ask students to find the locus of points \( P \) where \( \mathcal{P}_1(P) = \mathcal{P}_2(P) \).
Observation: Students should note that \( x^2 + y^2 \) terms cancel, proving the locus is a line. This is a powerful lesson in how "subtracting" two curved objects can yield a straight line.
2
Orthogonality Proof (15 mins)
Challenge students to prove that the radical axis is perpendicular to the line of centers.
Hint: Use coordinate geometry. Set the line of centers as the x-axis. The centers will be \( (x_1, 0) \) and \( (x_2, 0) \). The axis equation will then be of the form \( x = \text{constant} \).
3
Common Chords & Tangents (15 mins)
Discuss the radical axis for intersecting and tangent circles.
Intersecting: The power of any point on the common chord relative to either circle is the product of segments on that chord. Since the chord is shared, the power must be equal.Disjoint: The radical axis lies "between" them but closer to the smaller circle if the radii differ.
Discussion Point
"Can two concentric circles have a radical axis?" (Answer: No, the equations would yield a contradiction \( c_1 - c_2 = 0 \), meaning the 'line' is at infinity).
Construction Tip
To construct the axis for disjoint circles: Draw an auxiliary circle that intersects both. Draw the two common chords. Their intersection is a point on the radical axis.
Metric Relationships Circle Power // Sequence 1, Lesson 3 Guide
Radical Axis Map Worksheet Radical Axis Mapping
Loci of Equal Power
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Section 1: The Equation of Balance
Consider two circles:
1. Determine the Equation of the Radical Axis:
Set the power of a general point \( (x,y) \) equal for both circles and simplify the resulting expression.
Algebraic Work Area
Result:
Section 2: Locus and Orthogonality
2. The Centers and the Axis:
Find the centers \( O_1 \) and \( O_2 \) of the circles in Section 1. Show that the line segment \( O_1O_2 \) is perpendicular to the radical axis you found in Question 1.
3. Common Chord Property:
If two circles intersect at points \( M \) and \( N \), what is the power of point \( M \) relative to both circles? Use this to explain why the radical axis must pass through \( M \) and \( N \).
Section 3: Tangent Invariance
"For any point on the radical axis outside the circles, the tangents drawn to either circle have the same length."
4. Design a Construction:
Suppose you are given two disjoint circles. Describe a step-by-step process to find their radical axis using only a straightedge and compass, by finding points of equal power.
Metric Relationships Circle Power // Sequence 1, Lesson 3 Worksheet
Three Circle Nexus Slides The Radical Center
Nexus of Concurrency & Coaxial Systems
Lesson 04 // Undergrad Circle Power
Concurrency of Radical Axes
Given three circles with non-collinear centers...
"The three radical axes (one for each pair of circles) are concurrent at a single point."
This point is the Radical Center.
RC
Why Concurrent?
1
Let P be the intersection of radical axes \( L_{12} \) and \( L_{23} \).
2
Since \( P \in L_{12} \), we have \( \mathcal{P}_1(P) = \mathcal{P}_2(P) \).
3
Transitive Property: \( \mathcal{P}_1(P) = \mathcal{P}_3(P) \).
Coaxial Systems
A Coaxial System of circles is a family where every pair shares the same radical axis.
Algebraic Form
\( C_1 + \lambda C_2 = 0 \)
As \( \lambda \) varies, we generate a family of circles. Any circle in this family shares a radical axis with any other.
Types of Pencils:
Elliptic: Intersecting (common points)
Parabolic: Tangent (limiting case)
Hyperbolic: Disjoint (limiting points)
Application: Apollonian Gaskets
The radical center is the center of the unique circle that intersects all three given circles orthogonally.
Radius of orthogonal circle = \( \sqrt{\mathcal{P}_{RC}} \)
Nexus Navigation Teacher Guide Nexus Navigation Guide
Metric Relationships // Lesson 04: Radical Center & Coaxial Systems
UG-GEO-04
Lesson Purpose
This lesson explores the concurrency of three radical axes and introduces coaxial systems (circle pencils). The goal is for students to solve advanced construction problems by leveraging the existence of a unique point of equal power for three circles.
Theoretical Pillars
Concurrency Proof (Transitive Property)
Orthogonal Intersections
Linear Combinations of Circles
Instructional Steps
1
Proving Concurrency (15 mins)
Guide students through the logic: if P has equal power to circles 1 and 2, and circles 2 and 3, it must have equal power to 1 and 3.
Nuance: If the centers are collinear, the radical axes are parallel (intersecting at infinity). If non-collinear, they meet at a finite point.
2
Defining the Pencil (20 mins)
Introduce the family \( C_1 + \lambda C_2 = 0 \).
Elliptic: All circles pass through two common points (the intersections).Hyperbolic: Non-intersecting circles. Discuss "Limiting Points" (circles with radius 0).
3
The Orthogonal Circle (15 mins)
Explain that a circle centered at the Radical Center with radius \( R = \sqrt{\mathcal{P}_{RC}} \) will intersect all three circles at 90 degrees.
Application: This is a key step in solving Apollonian construction problems (finding circles tangent to three given circles).
Case Study
What happens if two circles are concentric? (The radical axis is at infinity). What if all three centers are collinear? (The axes are parallel, the Radical Center is a point at infinity).
Advanced Proof
Ask students to prove that the Radical Center of three circles is the orthocenter of some triangle if the circles are centered at the vertices and satisfy certain radius conditions.
Metric Relationships Circle Power // Sequence 1, Lesson 4 Guide
Radical Center Challenge Worksheet Radical Center Nexus
Concurrency and Family of Circles
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Part 1: The Three-Circle Theorem
Let \( C_1, C_2, C_3 \) be three circles with non-collinear centers. Let \( L_{ij} \) be the radical axis of circle \( C_i \) and \( C_j \).
1. The Transitive Proof:
"Suppose point P lies on the intersection of \( L_{12} \) and \( L_{23} \). Using the definition of Power of a Point, prove that P must also lie on \( L_{13} \)."
Formal Step-by-Step Proof
Part 2: Circle Pencils
2. Algebraic Pencil Analysis:
Consider the family of circles \( (x^2 + y^2 - 4x) + \lambda(x^2 + y^2 - 4y) = 0 \). Find the equation of the radical axis shared by any two distinct circles in this family.
Algebraic Work Area
3. The Orthogonal Discovery:
The Radical Center \( RC \) of three circles is 10 units from circle \( C_1 \) (radius 6), 13 units from circle \( C_2 \) (radius 11), and 15 units from circle \( C_3 \) (radius 13.5). Calculate the power of \( RC \) relative to each circle and find the radius of the circle centered at \( RC \) that is orthogonal to all three.
Construction Logic
Suppose you have three disjoint circles. You wish to find the radical center. Describe how you would use an auxiliary circle to find the radical axis of the first two, and then find the radical center by repeating the process.
Metric Relationships Circle Power // Sequence 1, Lesson 4 Worksheet
Space Inversion Slides Space Inversion
Orthogonal Condition & Conformal Mappings
Lesson 05 // Undergrad Circle Power
Orthogonal Circles
Two circles are orthogonal if they intersect at a 90° angle (tangents are perpendicular).
The Metric Condition:
\[ d^2 = r_1^2 + r_2^2 \]
Pythagorean theorem applied to centers and intersection.
Key Link: \( \mathcal{P}_{\Gamma_1}(O_2) = r_2^2 \)
Defining Inversion
Inversion in a circle \( \Gamma \) with center O and radius r maps a point P to P' such that:
\[ OP \cdot OP' = r^2 \]
P' lies on the ray OP
Core Intuition:
Inside maps to Outside
Outside maps to Inside
Points on the circle are Fixed
Center O maps to Infinity
The Geometry of Inversion
Mapping Rules:
Line (not thro O) → Circle (thro O)
Circle (not thro O) → Another Circle
Line (thro O) → Maps to Itself
Conformal Property
Inversion preserves angles but reverses orientation.
This makes it a powerful tool for simplifying problems involving tangency and orthogonality.
The Power of Inversion
"Inversion allows us to transform a pair of non-intersecting circles into a pair of concentric circles, simplifying complex chain-of-circles problems into basic symmetry."
Preserves Tangency
Maps Orthogonal to Orthogonal
Unifies Metric Geometry
Geometric Mirror Teacher Guide Geometric Mirror Guide
Metric Relationships // Lesson 05: Orthogonal Circles & Inversion
UG-GEO-05
Lesson Purpose
This culminating lesson connects Power of a Point to transformational geometry. By studying orthogonal circles and the geometry of inversion, students will see how the scalar power invariant underpins conformal mappings.
Transformational Pillars
Orthogonality Condition (\( d^2 = r_1^2 + r_2^2 \))
Inversion definition (\( OP \cdot OP' = r^2 \))
Conformal preservation of angles
Instructional Steps
1
The Power of Orthogonality (15 mins)
Derive the condition \( d^2 = r_1^2 + r_2^2 \) by considering the right triangle formed by the centers and the intersection point.
"Point out that the Power of the center of one circle relative to the other is precisely the square of its own radius if the circles are orthogonal."
2
Introduction to Inversion (20 mins)
Define inversion algebraically. Emphasize that points on the "Circle of Inversion" are fixed. Use dynamic software to show how lines map to circles.
Key Property: If a circle is orthogonal to the circle of inversion, it maps to itself. This is a profound geometric link.
3
The Conformal Property (15 mins)
Discuss why inversion is "conformal." While distances are distorted, angles are preserved. This allows us to use inversion to simplify complex tangency problems (like Steiner Porisms or Apollonian problems).
Workshop Key Notes
Calculating Inversion:
If center is at origin, inversion of \( (x, y) \) in circle of radius \( R \) is:
Orthogonal Pair Construction:
To construct a circle orthogonal to \( \Gamma \): Pick point P outside \( \Gamma \). Tangent segment \( PT \). Circle with center P and radius \( PT \) is orthogonal to \( \Gamma \).
Metric Relationships Circle Power // Sequence 1, Lesson 5 Guide
The Inversion Lab Worksheet The Inversion Lab
Reflective Transformations and Conformality
Name:
Date:
Section 1: The Orthogonal Condition
1. Center-Radius Check:
Two circles have equations \( \Gamma_1: x^2 + y^2 = 25 \) and \( \Gamma_2: (x-12)^2 + (y-5)^2 = 144 \). Determine if these circles are orthogonal by checking the metric condition \( d^2 = r_1^2 + r_2^2 \).
Calculation Area
Section 2: Mapping Logic
2. Point Mapping:
Let circle \( \Gamma \) be centered at the origin with radius \( r = 4 \). Find the image of the following points under inversion in \( \Gamma \):
A: (0, 2)
B: (8, 0)
C: (4, 0)
3. Transformational Intuition:
Consider a circle \( \Sigma \) that passes through the origin \( O \). When we invert \( \Sigma \) in another circle centered at \( O \), the image is a straight line. Explain why the image "un-curves" based on the limit behavior of inversion as a point approaches the center.
The Culminating Connection
We have seen that Power of a Point is invariant for any chord through P. In inversion, the product \( OP \cdot OP' = r^2 \) is also an invariant. How does the definition of inversion essentially treat the power of point \( P \) relative to the circle of inversion?
Metric Relationships Circle Power // Sequence 1, Lesson 5 Worksheet
Circle Power Answer Key Unit Answer Key
Metric Relationships // Complete Sequence Solutions
KEY-MRCP
Lesson 1: Invariant Product
1. Power Calculation: \( P = (x_0^2 + y_0^2) - r^2 = (3^2 + (-2)^2) - 25 = (9 + 4) - 25 = 13 - 25 = -12 \).
2. Sign Interpretation: Power is negative (-12), therefore point P is inside the circle.
3. Horizontal Case: Center O(0,0), radius r. Point P(d,0). Intersections A(r,0) and B(-r,0). Distances from P: \( PA = |r - d| \), \( PB = |-r - d| \). Signed product: \( (r-d)(-r-d) = -(r-d)(r+d) = -(r^2 - d^2) = d^2 - r^2 \).
4. Tangent Relation: For an external point, \( \mathcal{P}(P) = d^2 - r^2 \). By Pythagorean theorem on triangle OPT (where T is tangent point), \( PT^2 + r^2 = d^2 \implies PT^2 = d^2 - r^2 \). Thus, \( \text{Power} = \text{Tangent Length}^2 \).
Lesson 2: Similarity Link
1. Proof: ∠PAC = ∠PDB (angles inscribed in arc CB). ∠APC = ∠DPB (vertical). By AA similarity, \( \Delta PAC \sim \Delta PDB \). Therefore \( PA/PD = PC/PB \implies PA \cdot PB = PC \cdot PD \).
2. Continuity: As a secant rotates, the two intersection points A and B move toward each other. At the tangent limit, A=B=T. The product \( PA \cdot PB \) becomes \( PT \cdot PT = PT^2 \).
3. Application: Power = 64. Interior chord segment = 12. Let PA = x. Then PB = x + 12 (external point) is incorrect; for an interior point, the power is negative. If the power is stated as "64", it usually refers to the absolute value for an interior point. If P is external, \( x(x+12) = 64 \implies x^2 + 12x - 64 = 0 \implies (x+16)(x-4) = 0 \implies x = 4 \).
Lesson 3: Radical Axis Map
1. Radical Axis: \( (x^2 + y^2 - 4x - 6y + 9) = (x^2 + y^2 + 2x + 4y - 11) \). Canceling terms: \( -4x - 6y + 9 = 2x + 4y - 11 \implies 6x + 10y - 20 = 0 \implies 3x + 5y - 10 = 0 \).
2. Orthogonality: \( C_1(2, 3) \), \( C_2(-1, -2) \). Slope of centers \( m = \frac{-2-3}{-1-2} = \frac{-5}{-3} = 5/3 \). Slope of radical axis (\( 3x + 5y = 10 \)) is \( -3/5 \). Since \( (5/3) \cdot (-3/5) = -1 \), they are perpendicular.
Lesson 4: Radical Center Nexus
2. Pencil Axis: \( (1-\lambda)x^2 + (1-\lambda)y^2 - 4x + 4\lambda y = 0 \). To find the radical axis of the family, subtract two members (say \( \lambda=0 \) and \( \lambda=1 \)): \( (x^2 + y^2 - 4x) - (x^2 + y^2 - 4y) = 0 \implies -4x + 4y = 0 \implies y = x \).
3. Orthogonal Circle: Power \( \mathcal{P}_1 = 10^2 - 6^2 = 100 - 36 = 64 \). \( \mathcal{P}_2 = 13^2 - 11^2 = 169 - 121 = 48 \). This means the point provided is NOT the Radical Center. Radius of orthogonal circle = \( \sqrt{\text{Power}} \).
