Question
Verify that the equation for the line tangent to the circle (x-3)^2 + (y+1)^2 = 169 at the point (15,4) is indeed y = (-12/5)x + 40
Answers
GPT-4o mini
To verify that the equation of the tangent line to the circle given by \((x - 3)^2 + (y + 1)^2 = 169\) at the point \((15,4)\) is \(y = \frac{-12}{5}x + 40\), we'll follow these steps:
1. **Find the center and radius of the circle:**
The equation of the circle is in the standard form \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.
From the equation \((x - 3)^2 + (y + 1)^2 = 169\):
- Center \((h, k) = (3, -1)\)
- Radius \(r = \sqrt{169} = 13\)
2. **Find the slope of the radius to the tangent point (15, 4):**
The slope of the radius from the center \((3, -1)\) to the point \((15, 4)\) can be calculated as follows:
\[
m_{\text{radius}} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - (-1)}{15 - 3} = \frac{4 + 1}{15 - 3} = \frac{5}{12}
\]
3. **Find the slope of the tangent line:**
The slope of the tangent line is the negative reciprocal of the slope of the radius:
\[
m_{\text{tangent}} = -\frac{1}{m_{\text{radius}}} = -\frac{1}{\frac{5}{12}} = -\frac{12}{5}
\]
4. **Use the point-slope form to find the equation of the tangent line:**
Using the point-slope form of the equation of a line, \(y - y_1 = m(x - x_1)\):
Plugging in the point \((15, 4)\) and the slope \(-\frac{12}{5}\):
\[
y - 4 = -\frac{12}{5}(x - 15)
\]
Distributing the slope:
\[
y - 4 = -\frac{12}{5}x + \frac{12}{5} \times 15
\]
\[
y - 4 = -\frac{12}{5}x + \frac{180}{5}
\]
\[
y - 4 = -\frac{12}{5}x + 36
\]
Now, add 4 to each side:
\[
y = -\frac{12}{5}x + 36 + 4
\]
\[
y = -\frac{12}{5}x + 40
\]
This confirms that the equation for the line tangent to the circle at the point \((15, 4)\) is indeed:
\[
y = -\frac{12}{5}x + 40
\]
Thus, the given equation is verified.
1. **Find the center and radius of the circle:**
The equation of the circle is in the standard form \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.
From the equation \((x - 3)^2 + (y + 1)^2 = 169\):
- Center \((h, k) = (3, -1)\)
- Radius \(r = \sqrt{169} = 13\)
2. **Find the slope of the radius to the tangent point (15, 4):**
The slope of the radius from the center \((3, -1)\) to the point \((15, 4)\) can be calculated as follows:
\[
m_{\text{radius}} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - (-1)}{15 - 3} = \frac{4 + 1}{15 - 3} = \frac{5}{12}
\]
3. **Find the slope of the tangent line:**
The slope of the tangent line is the negative reciprocal of the slope of the radius:
\[
m_{\text{tangent}} = -\frac{1}{m_{\text{radius}}} = -\frac{1}{\frac{5}{12}} = -\frac{12}{5}
\]
4. **Use the point-slope form to find the equation of the tangent line:**
Using the point-slope form of the equation of a line, \(y - y_1 = m(x - x_1)\):
Plugging in the point \((15, 4)\) and the slope \(-\frac{12}{5}\):
\[
y - 4 = -\frac{12}{5}(x - 15)
\]
Distributing the slope:
\[
y - 4 = -\frac{12}{5}x + \frac{12}{5} \times 15
\]
\[
y - 4 = -\frac{12}{5}x + \frac{180}{5}
\]
\[
y - 4 = -\frac{12}{5}x + 36
\]
Now, add 4 to each side:
\[
y = -\frac{12}{5}x + 36 + 4
\]
\[
y = -\frac{12}{5}x + 40
\]
This confirms that the equation for the line tangent to the circle at the point \((15, 4)\) is indeed:
\[
y = -\frac{12}{5}x + 40
\]
Thus, the given equation is verified.