Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![## Finding the Equation of a Circle
### Problem Statement
**Find an equation of the circle that satisfies the given conditions:**
- **Center:** \((-2, 9)\)
- **Passes through:** \((-7, -6)\)
### Explanation
To find the equation of a circle, we use the standard form, which is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \((h, k)\) is the center of the circle and \(r\) is the radius.
Given:
- Center, \((h, k) = (-2, 9)\)
- A point on the circle, \((x_1, y_1) = (-7, -6)\)
First, we calculate the radius \(r\) using the distance formula between the center and the given point:
\[ r = \sqrt{(x_1 - h)^2 + (y_1 - k)^2} \]
Substitute the given values:
\[ r = \sqrt{(-7 - (-2))^2 + (-6 - 9)^2} \]
\[ r = \sqrt{(-7 + 2)^2 + (-6 - 9)^2} \]
\[ r = \sqrt{(-5)^2 + (-15)^2} \]
\[ r = \sqrt{25 + 225} \]
\[ r = \sqrt{250} \]
\[ r = 5\sqrt{10} \]
Now, we can write the equation of the circle using the standard form:
\[ (x - (-2))^2 + (y - 9)^2 = (5\sqrt{10})^2 \]
Simplifying,
\[ (x + 2)^2 + (y - 9)^2 = 250 \]
Therefore, the equation of the circle is:
\[ (x + 2)^2 + (y - 9)^2 = 250 \]
### Conclusion
This equation represents a circle with a center at \((-2, 9)\) and a radius of \(5\sqrt{10}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2c2bdfc9-f0b1-450f-8783-d3d092dc9ff9%2F98be59f3-ada2-4e33-b11a-d7b288d7bb61%2Fnxm5821a_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Finding the Equation of a Circle
### Problem Statement
**Find an equation of the circle that satisfies the given conditions:**
- **Center:** \((-2, 9)\)
- **Passes through:** \((-7, -6)\)
### Explanation
To find the equation of a circle, we use the standard form, which is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \((h, k)\) is the center of the circle and \(r\) is the radius.
Given:
- Center, \((h, k) = (-2, 9)\)
- A point on the circle, \((x_1, y_1) = (-7, -6)\)
First, we calculate the radius \(r\) using the distance formula between the center and the given point:
\[ r = \sqrt{(x_1 - h)^2 + (y_1 - k)^2} \]
Substitute the given values:
\[ r = \sqrt{(-7 - (-2))^2 + (-6 - 9)^2} \]
\[ r = \sqrt{(-7 + 2)^2 + (-6 - 9)^2} \]
\[ r = \sqrt{(-5)^2 + (-15)^2} \]
\[ r = \sqrt{25 + 225} \]
\[ r = \sqrt{250} \]
\[ r = 5\sqrt{10} \]
Now, we can write the equation of the circle using the standard form:
\[ (x - (-2))^2 + (y - 9)^2 = (5\sqrt{10})^2 \]
Simplifying,
\[ (x + 2)^2 + (y - 9)^2 = 250 \]
Therefore, the equation of the circle is:
\[ (x + 2)^2 + (y - 9)^2 = 250 \]
### Conclusion
This equation represents a circle with a center at \((-2, 9)\) and a radius of \(5\sqrt{10}\).
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