---### 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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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

Chapter7: Analytic Trigonometry

Section7.6: The Inverse Trigonometric Functions

Problem 93E

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Concept explainers

Coordinate Geometry

Coordinate geometry, also known as analytic geometry or Cartesian geometry in classical mathematics, is a type of geometry that is studied using a coordinate system. Coordinate geometry is a branch of mathematics that describes the position of points on a plane using an ordered pair of numbers called coordinates.

Question

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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

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