Write the equation of the circle centered at (2, - 1) that passes through (14, – 13).
Write the equation of the circle centered at (2, - 1) that passes through (14, – 13).
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE:
1. Give the measures of the complement and the supplement of an angle measuring 35°.
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![### Question 3
**Write the equation of the circle centered at (2, -1) that passes through (14, -13).**
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### Explanation:
To find the equation of a circle in the standard form \((x - h)^2 + (y - k)^2 = r^2\), we need the center \((h, k)\) and the radius \(r\).
Given:
- The center of the circle \((h, k) = (2, -1)\)
- A point on the circle \((x_1, y_1) = (14, -13)\)
First, calculate the radius \(r\) as the distance between the center and the given point using the distance formula:
\[ r = \sqrt{(x_1 - h)^2 + (y_1 - k)^2} \]
Substitute the coordinates:
\[ r = \sqrt{(14 - 2)^2 + (-13 + 1)^2} \]
\[ r = \sqrt{12^2 + (-12)^2} \]
\[ r = \sqrt{144 + 144} \]
\[ r = \sqrt{288} \]
\[ r = 12\sqrt{2} \]
Now, substitute \(h\), \(k\), and \(r\) into the standard form equation:
\[ (x - 2)^2 + (y + 1)^2 = (12\sqrt{2})^2 \]
\[ (x - 2)^2 + (y + 1)^2 = 288 \]
Thus, the equation of the circle is:
\[ (x - 2)^2 + (y + 1)^2 = 288 \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2e33d8a8-4a77-428f-baec-aa6f37a9e0af%2F7b01c22a-528a-4e03-a059-8aa50360d4e9%2Fmo6azhj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question 3
**Write the equation of the circle centered at (2, -1) that passes through (14, -13).**
---
**Submit your question below for help:**
[Submit Question]
or refer to the provided video tutorial for further assistance:
[Video]
---
### Explanation:
To find the equation of a circle in the standard form \((x - h)^2 + (y - k)^2 = r^2\), we need the center \((h, k)\) and the radius \(r\).
Given:
- The center of the circle \((h, k) = (2, -1)\)
- A point on the circle \((x_1, y_1) = (14, -13)\)
First, calculate the radius \(r\) as the distance between the center and the given point using the distance formula:
\[ r = \sqrt{(x_1 - h)^2 + (y_1 - k)^2} \]
Substitute the coordinates:
\[ r = \sqrt{(14 - 2)^2 + (-13 + 1)^2} \]
\[ r = \sqrt{12^2 + (-12)^2} \]
\[ r = \sqrt{144 + 144} \]
\[ r = \sqrt{288} \]
\[ r = 12\sqrt{2} \]
Now, substitute \(h\), \(k\), and \(r\) into the standard form equation:
\[ (x - 2)^2 + (y + 1)^2 = (12\sqrt{2})^2 \]
\[ (x - 2)^2 + (y + 1)^2 = 288 \]
Thus, the equation of the circle is:
\[ (x - 2)^2 + (y + 1)^2 = 288 \]
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