/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 94 Find an equation of the circle t... [FREE SOLUTION] | 91Ó°ÊÓ

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Find an equation of the circle that satisfies the given conditions. Endpoints of a diameter are \(P(-1,3)\) and \(Q(7,-5)\)

Short Answer

Expert verified
The equation of the circle is \((x - 3)^2 + (y + 1)^2 = 32\).

Step by step solution

01

Calculate the center of the circle

The center of the circle is the midpoint of the diameter. Use the midpoint formula for points \(P(x_1, y_1)\) and \(Q(x_2, y_2)\): \(\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\). Here, \(x_1 = -1\), \(y_1 = 3\), \(x_2 = 7\), \(y_2 = -5\). Calculate:\[\left( \frac{-1 + 7}{2}, \frac{3 + (-5)}{2} \right) = \left( \frac{6}{2}, \frac{-2}{2} \right) = (3, -1)\]. The center of the circle is \((3, -1)\).
02

Calculate the radius of the circle

The radius is half the distance of the diameter. First, find the distance between points \(P(-1,3)\) and \(Q(7,-5)\) using the distance formula: \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). Calculate:\[\sqrt{(7 - (-1))^2 + (-5 - 3)^2} = \sqrt{(7 + 1)^2 + (-8)^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2}\].The radius is half of this distance: \(\frac{8\sqrt{2}}{2} = 4\sqrt{2}\).
03

Write the equation of the circle

The general equation of a circle with center \((h, k)\) and radius \(r\) is \((x - h)^2 + (y - k)^2 = r^2\). Substitute \(h = 3\), \(k = -1\), and \(r = 4\sqrt{2}\):\[(x - 3)^2 + (y + 1)^2 = (4\sqrt{2})^2 = 32\].Thus, the equation is \((x - 3)^2 + (y + 1)^2 = 32\).

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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 handy tool in geometry that helps us find the center of a line segment. Given two points, the formula finds their average and provides the point directly in the middle. If you have points \(P(x_1, y_1)\) and \(Q(x_2, y_2)\), the midpoint \(M\) is calculated as:
  • \(\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\)
In the context of a circle, especially when dealing with a diameter, this point becomes significant. The midpoint of a diameter serves as the circle's center.
For example, if the endpoints of a diameter are \((-1, 3)\) and \((7, -5)\), computing the midpoint gives us the center at \((3, -1)\). This practical application of the midpoint formula allows us to easily locate the center, an essential step in defining a circle's equation.
Distance Formula
The distance formula is used to determine how far apart two points are in a plane. This is crucial in calculating the length of the diameter of a circle, which then helps in finding the circle's radius. Derived from the Pythagorean theorem, the distance between two points \(P(x_1, y_1)\) and \(Q(x_2, y_2)\) is:
  • \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
Applying this to points \((-1,3)\) and \((7,-5)\), we find the distance to be \(8\sqrt{2}\). This length represents the diameter of the circle.
To find the radius, simply halve this distance, which provides \(4\sqrt{2}\). The distance formula thus provides the necessary value of the radius, allowing us to express other circle properties.
General Equation of a Circle
The general equation of a circle ties all information about the circle into one neat expression. A circle with center \((h, k)\) and radius \(r\) is modeled by:
  • \((x - h)^2 + (y - k)^2 = r^2\)
This concise equation embodies the circle’s defining characteristics.
The center \((3, -1)\) and radius \(4\sqrt{2}\) yield the equation \((x - 3)^2 + (y + 1)^2 = 32\). Here, the transformations \((x - h)\) and \((y - k)\) adjust the general form to match the center location while \(r^2\) adjusts the size of the circle.
Understanding this equation is an essential skill. It allows you to effectively describe the size and location of a circle on the coordinate plane.

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