Problem 11 - The Calculus with Analytic Geometry

Question

Given the circle having the equation \(x^{2}+y^{2}=9\), find (a) the shortest distance from the point \((4,5)\) to a point on the circle, and (b) the longest distance from the point \((4,5)\) to a point on the circle.

Step-by-Step Solution

Answer

Shortest distance: 3.40. Longest distance: 9.40.

Step 1: Identify the center and radius of the circle

The equation of the circle is given by \(x^2 + y^2 = 9\). This is in the standard form \(x^2 + y^2 = r^2\), where the center is \( (0,0) \) and the radius \( r = 3 \).

Step 2: Calculate the distance from the center of the circle to the point (4,5)

The distance formula between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \].Substitute the given points \( (0,0) \) and \( (4,5) \): \[ d = \sqrt{(4 - 0)^2 + (5 - 0)^2} = \sqrt{16 + 25} = \sqrt{41} \approx 6.40 \]

Step 3: Determine the shortest distance from the point (4,5) to the circle

The shortest distance from the point (4,5) to the circle is the distance from the point to the circle's circumference, which is \( d - r \). Using the computed distance \( d = \sqrt{41} \approx 6.40 \) and the radius \( r = 3 \), the shortest distance is \[ 6.40 - 3 = 3.40 \]

Step 4: Determine the longest distance from the point (4,5) to the circle

The longest distance from the point (4,5) to the circle is the distance from the point to the farthest point on the circumference, which is \( d + r \). Using the computed distance \( d = \sqrt{41} \approx 6.40 \) and the radius \( r = 3 \), the longest distance is \[ 6.40 + 3 = 9.40 \]

Key Concepts

Circle Equations

A circle can be described algebraically using its equation. The general form for the equation of a circle is \[ (x - h)^2 + (y - k)^2 = r^2 \]where

• \( (h,k) \) is the center of the circle
• \( r \) is the radius

In this exercise, the equation is given as \( x^2 + y^2 = 9 \). This equation matches the standard form if we consider \( (h,k) = (0,0) \). Hence, the center of this circle is at the origin.

The term \( r^2 \) in the equation represents the radius squared. Here, \( r^2 = 9 \), so the radius \( r = 3 \).

Distance Formula

The distance formula helps to find the distance between two points in a plane. It is given by \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]where

• \( (x_1,y_1) \) and \( (x_2,y_2) \) are the coordinates of the two points

In the problem, we need to find the distance between the point \( (4,5) \) and the center of the circle \( (0,0) \).

Substituting the values in the formula gives \[ d = \sqrt{(4 - 0)^2 + (5 - 0)^2} = \sqrt{16 + 25} = \sqrt{41} \approx 6.40 \]. Therefore, the distance from the point \( (4,5) \) to the circle's center is approximately 6.40 units.

Radius in Geometry

In geometry, the radius of a circle is the distance from the center of the circle to any point on its circumference. For the equation of a circle \( x^2 + y^2 = 9 \), we have determined that the radius \( r \) is 3.

The radius plays a crucial role in determining distances from external points to the circle. To find the shortest and longest distances from a point outside the circle to the circle itself, you use the center-to-point distance and adjust by the radius:

• Shortest distance = center-to-point distance - radius
• Longest distance = center-to-point distance + radius

With our calculations, we found:
Shortest distance = \( 6.40 - 3 = 3.40 \)
Longest distance = \( 6.40 + 3 = 9.40 \)

Points on a Circle

A point lies on a circle if its coordinates satisfy the circle's equation. For the circle given by \( x^2 + y^2 = 9 \), any point \( (x,y) \) on the circle will make the equation true.

When looking for points on the circle from another external point, geometric properties are utilized:

• The shortest distance to the circle is found by moving directly towards the nearest point on its edge
• The longest distance is by moving directly away to the farthest point on the edge

Understanding these concepts can simplify problems involving distances from points to circles and enhance your grasp of geometric relationships.