Problem 52 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equation of the circle circumscribed about the right triangle whose vertices are \((0,0),(8,0)\), and \((0,6)\).

Step-by-Step Solution

Answer

The equation of the circumscribed circle is \((x-4)^2 + (y-3)^2 = 25\).

Step 1: Understand the Points and Triangle

The vertices \( (0,0), (8,0), (0,6) \) describe a right triangle. The base of the triangle lies along the x-axis from \( (0,0) \) to \( (8,0) \), and the height extends vertically to \( (0,6) \). The right angle is at the origin \( (0,0) \).

Step 2: Calculate the Hypotenuse Midpoint

The hypotenuse is the line between the points \( (8,0) \) and \( (0,6) \). First, find its midpoint using the formula: \[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right). \] Thus, the midpoint is: \[ \left( \frac{8+0}{2}, \frac{0+6}{2} \right) = (4, 3). \]

Step 3: Calculate the Radius of the Circle

The radius of the circle is half the hypotenuse length since the circumcircle of a right triangle has its center at the midpoint of the hypotenuse. Use the distance formula for the hypotenuse length: \[ \sqrt{(8-0)^2 + (0-6)^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = 10. \] Therefore, the radius is \( 5 \) because it is half of 10.

Step 4: Formulate the Equation of the Circle

The general equation of a circle with center \( (h, k) \) and radius \( r \) is \( (x-h)^2 + (y-k)^2 = r^2 \). Here, the center is \( (4,3) \) and the radius is 5. Substitute these values into the equation: \[ (x-4)^2 + (y-3)^2 = 5^2, \] which simplifies to: \[ (x-4)^2 + (y-3)^2 = 25. \]

Key Concepts

Right Triangle

In geometry, a right triangle is a type of triangle that has one angle measuring 90 degrees. This angle is known as the right angle. In the given problem, the right angle is at the origin (0,0).
When dealing with right triangles, it's important to remember that the sides opposite the right angle are referred to as the base and the height, and the side opposite the right angle is called the hypotenuse.
The unique property of the right triangle allows us to work with the Pythagorean Theorem, which states:

• \[ a^2 + b^2 = c^2 \], where \( a \) and \( b \) are the lengths of the two legs, and \( c \) is the length of the hypotenuse.

In this problem, we have a triangle with the vertices (0,0), (8,0), and (0,6). The two legs of this triangle lie along the x and y axes, with lengths 8 and 6, respectively, which are very straightforward to identify since they align with the axes. This makes our calculations more convenient.

Circle Equation

A circle equation is used to represent the set of all points that are a fixed distance from a given point, the center. The general formula for a circle's equation in a Cartesian plane is:

• \( (x - h)^2 + (y - k)^2 = r^2 \)

Here, \( (h, k) \) represents the center of the circle, while \( r \) represents the radius. Understanding how to manipulate this equation is crucial in solving problems involving circles.
With the right triangle in this exercise, once you determine the center and radius of the circle—as calculated from its geometric properties—you can plug it directly into this formula. For our problem, we found the circle's center at (4,3), and its radius to be 5.
Plugging these values into the circle equation gives:

• \[ (x - 4)^2 + (y - 3)^2 = 25 \]

This is a powerful way to represent any circle in a 2-dimensional space once its geometric properties are known.

Midpoint Formula

The midpoint formula serves to find the point that is exactly halfway between two given points in the coordinate plane.
Mathematically, the formula is:

• \[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\]

This formula is straightforward and simply averages the x-coordinates and the y-coordinates of the given points.
In the context of this exercise, it helps to find the midpoint of the hypotenuse of the right triangle. Why is the midpoint important here? Because, for a right triangle circumscribed circle, the center of the circle is the midpoint of the hypotenuse.
This is an important property, valid for any right triangle. Calculating for the points (8,0) and (0,6), we found the midpoint to be (4,3), placing the center of our circumcircle elegantly here.

Distance Formula

The distance formula is a key concept that calculates the distance between two points in the plane, providing a straightforward way to measure length in 2D space. This distance can be found using the formula:

• \[ sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

In our exercise, this formula is used to find the length of the hypotenuse, which is necessary to determine the radius of the circumcircle. Remember, the circumcircle (or circumscribed circle) is the circle that passes through all the vertices of the triangle.
For the vertices (8,0) and (0,6), plugging into the formula gives a hypotenuse length of 10. Since the radius of the circle is half of the hypotenuse, we computed the radius to be 5.
This computation not only confirms the position and size of the circle but also demonstrates the utility of the distance formula in connecting geometric properties to algebraic expressions.