# Circle Inversion: Complete Study Guide for Geometric Inversion
## A Comprehensive Guide to Inversion with Respect to Circles

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## 🎯 Introduction: Understanding Circle Inversion

Welcome to the complete guide on **Circle Inversion** (also known as **Geometric Inversion**). Circle inversion is a powerful transformation in geometry that maps points inside a circle to points outside the circle and vice versa, while preserving certain geometric properties.

This transformation has fascinating applications in:
- Complex analysis and conformal mapping
- Geometric constructions and proofs
- Non-Euclidean geometry
- Physics and engineering problems
- Advanced mathematics competitions

**Key Learning Objective**: Master the concept of inversion, understand the fundamental inversion formula, and solve complex problems involving inverse points with respect to circles. 📐

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## 🔑 Fundamental Concepts of Circle Inversion

### What is Circle Inversion?

**Circle Inversion** is a transformation that maps each point P in the plane (except the center) to a corresponding point P' such that:
- The center O, point P, and point P' are **collinear**
- The relationship **OP × OP' = r²** holds (where r is the radius of the circle of inversion)

### Basic Properties of Inversion

1. **Reciprocal Relationship**: If P' is the inverse of P, then P is also the inverse of P'
2. **Points on the Circle**: Points on the circle of inversion map to themselves
3. **Inside ↔ Outside**: Points inside the circle map to points outside, and vice versa
4. **Center Behavior**: The center of inversion has no inverse (maps to infinity)

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## 📐 Essential Inversion Formulas

### 1. For Circle Centered at Origin

**Circle of Inversion**: $x^2 + y^2 = r^2$

**Inversion Formula**:
$$P'(x', y') = \left(\frac{r^2 x}{x^2 + y^2}, \frac{r^2 y}{x^2 + y^2}\right)$$

**Key Properties**:
- Distance relationship: $|OP| \times |OP'| = r^2$
- If $|OP| = d$, then $|OP'| = \frac{r^2}{d}$

### 2. For Circle Centered at Point (h, k)

**Circle of Inversion**: $(x - h)^2 + (y - k)^2 = r^2$

**Inversion Formula**:
$$P'(x', y') = \left(\frac{r^2(x - h)}{(x - h)^2 + (y - k)^2} + h, \frac{r^2(y - k)}{(x - h)^2 + (y - k)^2} + k\right)$$

**Key Properties**:
- The center (h, k) is the center of inversion
- Distance relationship: $|CP| \times |CP'| = r^2$ where C is the center

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## 📚 Worked Examples and Solutions

### Problem 1: Understanding the Basic Inversion Relationship
**Question:** In the figure, O is centre of circle, OP = 4 units, OQ = 8 units and OP' = 16 units, write the relation between P and P'.

![Circle inversion diagram](https://seekoguide.com/images/inversion/1.svg)

**Complete Solution:**

**Given Information:**
- OP = 4 units
- OQ = 8 units (radius of circle)
- OP' = 16 units

**Step 1: Check the inversion relationship**
For inversion with respect to a circle, we must have: $OP \times OP' = r^2$

**Step 2: Verify the relationship**
- $OP \times OP' = 4 \times 16 = 64$
- $r^2 = OQ^2 = 8^2 = 64$
- Since $64 = 64$, the relationship holds!

**Step 3: State the conclusion**
The relationship is: $OP \times OP' = r^2$

**Final Answer:** P' is the **inversion point of P** (and vice versa) with respect to the circle centered at O with radius 8 units.

### Problem 2: Finding Inverse with Respect to Circle Centered at Origin
**Question:** Find the inverse image of the point (4, 5) with respect to the circle $x^2 + y^2 = 10$.

**Complete Solution:**

**Given Information:**
- Circle: $x^2 + y^2 = 10$
- Center of circle: (0, 0)
- Radius: $r^2 = 10$ (so $r = \sqrt{10}$)
- Object point P(x, y) = (4, 5)
- Inversion point P'(x', y') = ?

**Step 1: Apply the inversion formula**
For circle centered at origin:
$$P'(x', y') = \left(\frac{r^2 x}{x^2 + y^2}, \frac{r^2 y}{x^2 + y^2}\right)$$

**Step 2: Substitute the values**
$$P'(x', y') = \left(\frac{10 \times 4}{4^2 + 5^2}, \frac{10 \times 5}{4^2 + 5^2}\right)$$

**Step 3: Calculate the denominator**
$4^2 + 5^2 = 16 + 25 = 41$

**Step 4: Complete the calculation**
$$P'(x', y') = \left(\frac{40}{41}, \frac{50}{41}\right)$$

**Final Answer:** The inverse of point (4, 5) is $\left(\frac{40}{41}, \frac{50}{41}\right)$

**Verification:** $OP \times OP' = \sqrt{41} \times \frac{\sqrt{40^2 + 50^2}}{41} = \sqrt{41} \times \frac{\sqrt{4100}}{41} = \sqrt{41} \times \frac{\sqrt{41 \times 100}}{41} = 10$ ✓

### Problem 3: Finding Inverse with Respect to Circle with Different Center
**Question:** Find the inverse image of the point (3, 4) about the circle $(x - 2)^2 + (y - 2)^2 = 36$.

**Complete Solution:**

**Given Information:**
- Circle: $(x - 2)^2 + (y - 2)^2 = 36$
- Center of circle (h, k) = (2, 2)
- Radius: r = 6 (since $r^2 = 36$)
- Object point (x, y) = (3, 4)
- Inversion point (x', y') = ?

**Step 1: Apply the inversion formula for non-origin center**
$$P'(x', y') = \left(\frac{r^2(x - h)}{(x - h)^2 + (y - k)^2} + h, \frac{r^2(y - k)}{(x - h)^2 + (y - k)^2} + k\right)$$

**Step 2: Calculate the distance from point to center**
- $(x - h) = 3 - 2 = 1$
- $(y - k) = 4 - 2 = 2$
- $(x - h)^2 + (y - k)^2 = 1^2 + 2^2 = 1 + 4 = 5$

**Step 3: Substitute into the formula**
$$P'(x', y') = \left(\frac{36 \times 1}{5} + 2, \frac{36 \times 2}{5} + 2\right)$$

**Step 4: Simplify**
$$P'(x', y') = \left(\frac{36}{5} + 2, \frac{72}{5} + 2\right)$$
$$P'(x', y') = \left(\frac{36 + 10}{5}, \frac{72 + 10}{5}\right)$$
$$P'(x', y') = \left(\frac{46}{5}, \frac{82}{5}\right)$$

**Final Answer:** The inverse of point (3, 4) about the given circle is $\left(\frac{46}{5}, \frac{82}{5}\right)$

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## 🧮 Practice Problems with Solutions

### Practice Set A: Basic Inversion Problems

**Problem A1:** Find the inverse of point (6, 8) with respect to the circle $x^2 + y^2 = 25$.

**Solution:**
- Center: (0, 0), $r^2 = 25$
- $x^2 + y^2 = 6^2 + 8^2 = 36 + 64 = 100$
- $P'(x', y') = \left(\frac{25 \times 6}{100}, \frac{25 \times 8}{100}\right) = \left(\frac{3}{2}, 2\right)$

**Answer: $\left(\frac{3}{2}, 2\right)$**

**Problem A2:** If P(2, 3) and P'(a, b) are inverse points with respect to circle $x^2 + y^2 = 26$, find a and b.

**Solution:**
- $r^2 = 26$, $x^2 + y^2 = 4 + 9 = 13$
- $a = \frac{26 \times 2}{13} = 4$
- $b = \frac{26 \times 3}{13} = 6$

**Answer: a = 4, b = 6**

### Practice Set B: Advanced Inversion Problems

**Problem B1:** Find the inverse of point (5, 1) with respect to circle $(x - 3)^2 + (y - 2)^2 = 16$.

**Solution:**
- Center: (3, 2), $r^2 = 16$
- $(x - h) = 5 - 3 = 2$, $(y - k) = 1 - 2 = -1$
- $(x - h)^2 + (y - k)^2 = 4 + 1 = 5$
- $x' = \frac{16 \times 2}{5} + 3 = \frac{32}{5} + 3 = \frac{47}{5}$
- $y' = \frac{16 \times (-1)}{5} + 2 = -\frac{16}{5} + 2 = -\frac{6}{5}$

**Answer: $\left(\frac{47}{5}, -\frac{6}{5}\right)$**

**Problem B2:** Show that if two points are inverse with respect to a circle, then the circle is the perpendicular bisector of the line segment joining the two points when extended.

**Solution:**
This is a theorem in inversion geometry. The proof involves showing that any circle orthogonal to the circle of inversion passes through both P and P', making the inversion circle the locus of centers of all such circles.

### Practice Set C: Applications and Properties

**Problem C1:** Find the equation of the circle that passes through point (1, 2) and its inverse with respect to circle $x^2 + y^2 = 9$.

**Solution:**
**Step 1:** Find inverse of (1, 2)
- $P' = \left(\frac{9 \times 1}{5}, \frac{9 \times 2}{5}\right) = \left(\frac{9}{5}, \frac{18}{5}\right)$

**Step 2:** Any circle passing through both points and orthogonal to $x^2 + y^2 = 9$
- The required circle has the form: $x^2 + y^2 + Dx + Ey + F = 0$ where it's orthogonal to the inversion circle.

**Problem C2:** Prove that inversion preserves angles.

**Solution:**
This is a fundamental theorem: **Inversion is a conformal transformation**, meaning it preserves angles between curves at their intersection points.

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## 🎨 Special Properties of Inversion

### Geometric Properties Preserved Under Inversion

1. **Angle Preservation**: Angles between curves are preserved
2. **Circle/Line Mapping**: 
   - Circles through the center → Lines not through the center
   - Circles not through the center → Circles not through the center
   - Lines through the center → Lines through the center
   - Lines not through the center → Circles through the center

### Important Theorems

**Theorem 1**: The inverse of a line passing through the center of inversion is the line itself.

**Theorem 2**: The inverse of a line not passing through the center of inversion is a circle passing through the center of inversion.

**Theorem 3**: The inverse of a circle passing through the center of inversion is a line not passing through the center.

**Theorem 4**: The inverse of a circle not passing through the center of inversion is another circle not passing through the center.

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## 💡 Problem-Solving Strategies

### Strategy 1: Direct Formula Application
1. **Identify** the circle of inversion (center and radius)
2. **Determine** which formula to use (origin-centered or general)
3. **Substitute** coordinates carefully
4. **Simplify** step by step
5. **Verify** using the fundamental relationship

### Strategy 2: Geometric Approach
1. **Draw** the configuration
2. **Use** the property $OP \times OP' = r^2$
3. **Apply** similar triangles when appropriate
4. **Check** collinearity of O, P, and P'

### Strategy 3: Verification Method
1. **Calculate** distances OP and OP'
2. **Verify** that $OP \times OP' = r^2$
3. **Check** that O, P, P' are collinear

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## 📊 Common Mistakes to Avoid

### Mistake 1: Wrong Formula Selection
❌ **Wrong:** Using origin formula for non-origin centered circles
✅ **Correct:** Always check the center of the circle of inversion

### Mistake 2: Arithmetic Errors
❌ **Wrong:** Making calculation errors in fractions
✅ **Correct:** Double-check all arithmetic, especially with fractions

### Mistake 3: Sign Confusion
❌ **Wrong:** Missing negative signs in coordinates
✅ **Correct:** Pay careful attention to coordinate signs

### Mistake 4: Distance Calculation
❌ **Wrong:** Using wrong distance in non-origin cases
✅ **Correct:** Always calculate distance from the center of inversion

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## 🎯 Examination Tips

### Time-Saving Techniques
1. **Master both formulas** - origin and general center cases
2. **Practice fraction arithmetic** - most answers involve fractions
3. **Use verification** - check your answer with $OP \times OP' = r^2$
4. **Draw diagrams** when possible for geometric insight

### Common Exam Question Types
1. **Direct inversion**: Find inverse of given point
2. **Finding unknowns**: Given one point and its inverse, find missing coordinates
3. **Relationship verification**: Prove two points are inverse
4. **Applications**: Using inversion in geometric constructions

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## 🏆 Advanced Applications

### Applications in Mathematics
- **Complex Analysis**: Möbius transformations
- **Hyperbolic Geometry**: Poincaré disk model
- **Geometric Proofs**: Elegant solutions using inversion
- **Conformal Mapping**: Preserving angles in transformations

### Applications in Physics and Engineering
- **Electrostatics**: Method of images
- **Fluid Dynamics**: Flow around obstacles
- **Optics**: Lens design and ray tracing
- **Antenna Theory**: Radiation patterns

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## 📝 Summary Checklist

By the end of this guide, you should be able to:

✅ **Understand the concept** of circle inversion and its geometric meaning  
✅ **Apply inversion formulas** for both origin-centered and general circles  
✅ **Find inverse points** given any point and circle of inversion  
✅ **Verify inversion relationships** using the fundamental property $OP \times OP' = r^2$  
✅ **Work with fractional coordinates** accurately in inversion problems  
✅ **Understand geometric properties** preserved under inversion  
✅ **Solve complex inversion problems** involving multiple steps  
✅ **Apply inversion** to geometric constructions and proofs  
✅ **Recognize applications** of inversion in advanced mathematics  
✅ **Use verification methods** to check solutions  

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## 🎓 Final Words

Circle inversion is one of the most elegant and powerful transformations in geometry. It provides:

1. **Beautiful geometric insights** into the nature of circles and lines
2. **Powerful problem-solving tools** for complex geometric constructions
3. **Deep connections** to advanced areas of mathematics
4. **Practical applications** in physics and engineering

**Key to Success:**
- **Practice the formulas** until they become second nature
- **Understand the geometric meaning** behind the algebraic calculations  
- **Visualize the transformation** - points inside map outside and vice versa
- **Use verification** as a powerful check on your work

Remember: **Inversion turns the inside out and the outside in, but angles stay the same!** 

This transformation has fascinated mathematicians for centuries and continues to provide elegant solutions to seemingly difficult problems. Master it, and you'll have a powerful tool in your mathematical toolkit! 🚀

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*This guide provides comprehensive coverage of circle inversion with detailed worked examples, practice problems, and advanced applications. Use it as your complete reference for mastering geometric inversion.*
