5. Let C be a circle, and let P be a point not on the circle. Prove that the maximum and minimum distances from P to a point X on C occur when the line XP goes through the center of C. [Hint: Choose coordinate systems so that C is defined by x² + y? = r2 and P is a point (a, 0) on the x-axis with a ±r; use calculus to find the maximum and minimum for the square of the distance. Don't forget to pay attention to endpoints and places where a derivative might not exist.]
5. Let C be a circle, and let P be a point not on the circle. Prove that the maximum and minimum distances from P to a point X on C occur when the line XP goes through the center of C. [Hint: Choose coordinate systems so that C is defined by x² + y? = r2 and P is a point (a, 0) on the x-axis with a ±r; use calculus to find the maximum and minimum for the square of the distance. Don't forget to pay attention to endpoints and places where a derivative might not exist.]
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.3: Lines
Problem 8E
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### Problem 5: Maximum and Minimum Distances from a Point to a Circle
**Problem Statement:**
Let \( C \) be a circle, and let \( P \) be a point not on the circle. Prove that the maximum and minimum distances from \( P \) to a point \( X \) on \( C \) occur when the line \( XP \) goes through the center of \( C \).
**Hint:** Choose coordinate systems so that \( C \) is defined by \( x^2 + y^2 = r^2 \) and \( P \) is a point \( (a,0) \) on the x-axis with \( a \neq \pm r \); use calculus to find the maximum and minimum for the square of the distance. Don’t forget to pay attention to endpoints and places where a derivative might not exist.
---
### Explanation:
1. **Choose Coordinate System:**
- Define the circle \( C \) with the equation \( x^2 + y^2 = r^2 \).
- Let \( P \) be the point \( (a, 0) \) on the x-axis, where \( a \neq \pm r \).
2. **Distance Formula:**
- The square of the distance from \( P(a, 0) \) to any point \( X(x, y) \) on the circle is given by:
\[
D^2 = (x - a)^2 + y^2
\]
3. **Maximize/Minimize the Distance:**
- Substituting \( y^2 = r^2 - x^2 \) from the circle's equation:
\[
D^2 = (x - a)^2 + (r^2 - x^2)
\]
- Simplifying:
\[
D^2 = x^2 - 2ax + a^2 + r^2 - x^2 = a^2 - 2ax + r^2
\]
- To find the critical points, take the derivative of \( D^2 \) with respect to \( x \), and set it to zero:
\[
\frac{d(D^2)}{dx} = -2a = 0
\]
- Since this derivative does not have a variable part except \( -2a](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8b0185de-0645-4c2a-aea4-e046d61ab5cb%2Ffbb7e82b-1453-4750-82f1-faafe4553d7b%2Fjgepr88.jpeg&w=3840&q=75)
Transcribed Image Text:---
### Problem 5: Maximum and Minimum Distances from a Point to a Circle
**Problem Statement:**
Let \( C \) be a circle, and let \( P \) be a point not on the circle. Prove that the maximum and minimum distances from \( P \) to a point \( X \) on \( C \) occur when the line \( XP \) goes through the center of \( C \).
**Hint:** Choose coordinate systems so that \( C \) is defined by \( x^2 + y^2 = r^2 \) and \( P \) is a point \( (a,0) \) on the x-axis with \( a \neq \pm r \); use calculus to find the maximum and minimum for the square of the distance. Don’t forget to pay attention to endpoints and places where a derivative might not exist.
---
### Explanation:
1. **Choose Coordinate System:**
- Define the circle \( C \) with the equation \( x^2 + y^2 = r^2 \).
- Let \( P \) be the point \( (a, 0) \) on the x-axis, where \( a \neq \pm r \).
2. **Distance Formula:**
- The square of the distance from \( P(a, 0) \) to any point \( X(x, y) \) on the circle is given by:
\[
D^2 = (x - a)^2 + y^2
\]
3. **Maximize/Minimize the Distance:**
- Substituting \( y^2 = r^2 - x^2 \) from the circle's equation:
\[
D^2 = (x - a)^2 + (r^2 - x^2)
\]
- Simplifying:
\[
D^2 = x^2 - 2ax + a^2 + r^2 - x^2 = a^2 - 2ax + r^2
\]
- To find the critical points, take the derivative of \( D^2 \) with respect to \( x \), and set it to zero:
\[
\frac{d(D^2)}{dx} = -2a = 0
\]
- Since this derivative does not have a variable part except \( -2a
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