Problem 45 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 35-46: Find an equation of the circle that satisfies the stated conditions. $$ \text { Endpoints of a diameter } A(4,-3) \text { and } B(-2,7) $$

Step-by-Step Solution

Answer

The equation is \((x - 1)^2 + (y - 2)^2 = 34\).

Step 1: Understand the Circle's Equation

The standard equation of a circle is given by \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center of the circle and \(r\) is the radius. To find the center and radius, we need to use the given endpoints of a diameter.

Step 2: Find the Center of the Circle

For a circle, the center is the midpoint of its diameter. The midpoint of a line segment with endpoints \((x_1, y_1)\) and \((x_2, y_2)\) is \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\). Here, it is \(\left(\frac{4 + (-2)}{2}, \frac{-3 + 7}{2}\right) = (1, 2)\). Thus, the center \((h, k)\) is \((1, 2)\).

Step 3: Calculate the Radius

The radius is half of the length of the diameter. The length \(d\) of the diameter can be found using the distance formula between points \((x_1, y_1)\) and \((x_2, y_2)\): \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). Applying this, the diameter is \(\sqrt{(-2 - 4)^2 + (7 + 3)^2} = \sqrt{36 + 100} = \sqrt{136}\). The radius \(r\) is then \(\frac{\sqrt{136}}{2} = \sqrt{34}\).

Step 4: Write the Equation of the Circle

Insert the center \((h, k) = (1, 2)\) and the radius \(r = \sqrt{34}\) into the circle equation format: \((x - h)^2 + (y - k)^2 = r^2\). This results in \((x - 1)^2 + (y - 2)^2 = 34\).

Key Concepts

Midpoint Formula

When given two endpoints of a line segment, you can easily find the midpoint, which is essentially the center of the segment. This can be very helpful in problems involving circles because the midpoint of the diameter is actually the center of the circle. To find the midpoint between two points

• Use the formula: \[\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\]
• x and y values are averaged separately.
• This midpoint becomes the center \((h, k)\) of the circle.

In the problem exercise provided, the points A(4,-3) and B(-2,7) were used to calculate the midpoint (1, 2), which thereby determined the circle's center.

Distance Formula

The distance formula is a crucial tool when working with geometric figures, especially circles. It helps in finding the length of a line segment, like a circle's diameter, between two points. The formula is:

• \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]
• It calculates the straight-line distance between two points.
• This distance is essential for determining other properties like the radius.

For the exercise, substituting the given points A(4,-3) and B(-2,7) into the formula, we identified the diameter's length as \(\sqrt{136}\). This served as a stepping stone to find the circle's radius.

Radius Calculation

To establish the size of a circle, knowing the radius is fundamental. The radius is simply half the length of the circle's diameter.

• First, calculate the diameter's length using the distance formula.
• Then, divide this length by 2 to get the radius.
• In formula terms, if \(d\) is the diameter, then \(r = \frac{d}{2}\).

From the exercise, after establishing that the diameter length was \(\sqrt{136}\), the radius came out to \(\sqrt{34}\). This radius is a key component in deriving the circle’s equation.

Diameter Endpoints

The endpoints of a diameter are critical pieces of information when defining a circle. They allow us to find not only the circle's center but also the radius. Remember:

• Endpoints directly help in calculating the center via the midpoint formula.
• Also, they contribute to determining the length of the diameter with help from the distance formula.
• Both these measurements lead to the final circle equation.

In the problem, by knowing that A(4,-3) and B(-2,7) are diameter endpoints, we were able to effectively establish that the circle’s center is (1, 2) and with the radius being \(\sqrt{34}\), forming the equation \((x - 1)^2 + (y - 2)^2 = 34\).