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 Tangent Line to a Circle at a Given Point
To find the equation of the tangent line to the given circle at the specified point, we first need to understand the equation of the circle and the geometry involved.
#### Circle Equation
The given equation of the circle is:
\[
(x + 2)^2 + (y - 3)^2 = 37
\]
This represents a circle centered at \((-2, 3)\) with radius \(\sqrt{37}\).
#### Point of Tangency
The specific point given is:
\[
(-3, -3)
\]
#### Diagram Explanation
The diagram shows a circle centered at \((-2, 3)\) with a radius intercepting the given point \((-3, -3)\). The marked point \((-3, -3)\) lies on the circumference of the circle.
### Steps to Find the Tangent Line
1. **Calculate the Slope of the Radius:**
- The slope of the line connecting the center \((-2, 3)\) to the point of tangency \((-3, -3)\) is:
\[
m_{\text{radius}} = \frac{3 - (-3)}{-2 - (-3)} = \frac{6}{1} = 6
\]
2. **Slope of the Tangent Line:**
- The tangent line will be perpendicular to the radius at the point of tangency. The slope of the tangent line (\(m_{\text{tangent}}\)) is the negative reciprocal of the slope of the radius:
\[
m_{\text{tangent}} = -\frac{1}{6}
\]
3. **Equation of the Tangent Line:**
- Using the point-slope form of the linear equation: \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is \((-3, -3)\):
\[
y + 3 = -\frac{1}{6}(x + 3)
\]
- Simplifying to the slope-intercept form (\(y = mx + b\)):
\[
y + 3 = -\frac{1}{6}x - \frac{1}{2}
\]
\[
y = -\frac{1}{6}x - \frac{1}{2} -](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8d04d9e8-9cf4-4d61-b8ac-39958688637f%2F61138b13-5e9f-4c4b-89db-50ae147b8639%2Fk3xr2z7.jpeg&w=3840&q=75)
Transcribed Image Text:### Finding the Equation of a Tangent Line to a Circle at a Given Point
To find the equation of the tangent line to the given circle at the specified point, we first need to understand the equation of the circle and the geometry involved.
#### Circle Equation
The given equation of the circle is:
\[
(x + 2)^2 + (y - 3)^2 = 37
\]
This represents a circle centered at \((-2, 3)\) with radius \(\sqrt{37}\).
#### Point of Tangency
The specific point given is:
\[
(-3, -3)
\]
#### Diagram Explanation
The diagram shows a circle centered at \((-2, 3)\) with a radius intercepting the given point \((-3, -3)\). The marked point \((-3, -3)\) lies on the circumference of the circle.
### Steps to Find the Tangent Line
1. **Calculate the Slope of the Radius:**
- The slope of the line connecting the center \((-2, 3)\) to the point of tangency \((-3, -3)\) is:
\[
m_{\text{radius}} = \frac{3 - (-3)}{-2 - (-3)} = \frac{6}{1} = 6
\]
2. **Slope of the Tangent Line:**
- The tangent line will be perpendicular to the radius at the point of tangency. The slope of the tangent line (\(m_{\text{tangent}}\)) is the negative reciprocal of the slope of the radius:
\[
m_{\text{tangent}} = -\frac{1}{6}
\]
3. **Equation of the Tangent Line:**
- Using the point-slope form of the linear equation: \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is \((-3, -3)\):
\[
y + 3 = -\frac{1}{6}(x + 3)
\]
- Simplifying to the slope-intercept form (\(y = mx + b\)):
\[
y + 3 = -\frac{1}{6}x - \frac{1}{2}
\]
\[
y = -\frac{1}{6}x - \frac{1}{2} -
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