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Chapter 1: Problem 78

Find the equation of the circle. Endpoints of a diameter are (-3,5) and (7,-5).

Short Answer

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Question: Find the equation of a circle with endpoints of a diameter (-3,5) and (7,-5). Answer: The equation of the circle is \((x-2)^2 + y^2 = 50\).

Step by step solution

01

Find the center of the circle

To find the midpoint (which is the center of the circle) between the two endpoints of the diameter, use the midpoint formula: Midpoint = \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\) Using the given points (-3,5) and (7,-5): Midpoint = \(\left(\frac{-3+7}{2}, \frac{5-5}{2}\right) = (2,0)\) So, the center of the circle is (2,0).
02

Find the radius of the circle

To find the radius, we calculate the distance between the center of the circle and one of the endpoints using the distance formula: Distance = \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\) Using the center (2, 0) and one of the endpoints (-3, 5): Radius = \(\sqrt{(-3-2)^2 + (5-0)^2} = \sqrt{25 + 25} = \sqrt{50}\) So, the radius of the circle is \(\sqrt{50}\).
03

Write the equation of the circle

Now that we have the center and the radius, we can write the equation of the circle using the standard equation: \((x-a)^2 + (y-b)^2 = r^2\) Where (a, b) is the center of the circle and r is the radius. Plugging in the values we found for the center and the radius: \((x-2)^2 + (y-0)^2 = (\sqrt{50})^2\) Simplifying the equation: \((x-2)^2 + y^2 = 50\) This is the equation of the circle with endpoints of the diameter (-3,5) and (7,-5).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Midpoint Formula
The midpoint formula is a fundamental concept in geometry, especially when dealing with circles. It provides the means to find the exact center point between two given points. In the context of a circle, this midpoint is often the center of the circle itself. The formula is given by
\textbf{Midpoint} = \(\bigg(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\bigg)\)
.
When we have the endpoints of a diameter, like in our exercise with points (-3,5) and (7,-5), we can apply this formula to locate the center of the circle, which is crucial for further steps like determining the circle's radius or its equation.
Radius of a Circle
The radius of a circle is simply the distance from the center of the circle to any point on the circle. Knowing how to calculate the radius is essential, as the standard equation of a circle revolves around this value.
Calculating the radius often involves the distance formula, which helps us measure the length between two points in a coordinate plane. Once we find the center using the midpoint formula, we can pair it with any point on the circle, such as an endpoint of a diameter, to determine the radius.
Distance Formula
The distance formula is used to calculate the distance between two points in the x-y plane and is derived from the Pythagorean theorem. It is expressed as \[Distance = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\]
.
This formula is pivotal when we need to find the length of the radius in our circle equation problem. By substituting the coordinates of the center and one endpoint of the diameter, we establish the radius, which we square to input into the circle's equation. As seen in the solution, precise usage of this formula roots down to understanding the properties of the circle in conjunction with Euclidean distance.
Standard Equation of a Circle
The standard equation of a circle provides a way to represent all points of a circle in a coordinate plane. It鈥檚 written as \[(x-a)^2 + (y-b)^2 = r^2\]
where \( (a, b) \) are the coordinates of the center, and \( r \) is the radius.
Utilizing the center and radius determined from the midpoint and distance formulas, we can construct the specific equation that models our circle. In our example solution, by plugging in the center coordinate (2, 0) and the radius \( \sqrt{50} \), we end up with a clear, standard form equation that effectively defines our circle given its diameter endpoints.

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Most popular questions from this chapter

Find an equation for the line satisfying the given conditions. \(x\) -intercept 5 and \(y\) -intercept -5.

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