The equation (s) of circle (s) touching `12x - 5y = 7` at (1, 1) and having radius 13 is/are

Options

A. `x^(2) + y^(2) - 22x + 12y - 1 = 0`
B. `x^(2) + y^(2) + 22x - 12y - 12 = 0`
C. `x^(2) + y^(2) - 26x - 8y + 16 = 0`
D. `x^(2) + y^(2) -26x + 8y + 16 = 0`

Answer

To solve the problem of finding the equations of circles touching the line \(12x - 5y = 7\) at the point \((1, 1)\) and having a radius of \(13\), we can follow these steps:

### Step 1: Identify the line equation and point of tangency
The given line is:
\[ 12x - 5y - 7 = 0 \]
The point of tangency is \((1, 1)\).

### Step 2: Find the slope of the line
To find the slope of the line, we can rearrange the line equation into slope-intercept form \(y = mx + b\):
\[ 5y = 12x - 7 \implies y = \frac{12}{5}x - \frac{7}{5} \]
The slope \(m\) of the line is \(\frac{12}{5}\).

### Step 3: Find the slope of the radius
The radius of the circle at the point of tangency is perpendicular to the tangent line. The slope of the radius \(m_r\) is the negative reciprocal of the slope of the line:
\[ m_r = -\frac{5}{12} \]

### Step 4: Write the equation of the radius
Using the point-slope form of the equation of a line, we can write the equation of the radius from the point \((1, 1)\):
\[ y - 1 = -\frac{5}{12}(x - 1) \]
Simplifying this, we get:
\[ y - 1 = -\frac{5}{12}x + \frac{5}{12} \implies y = -\frac{5}{12}x + \frac{17}{12} \]

### Step 5: Find the center of the circle
Let the center of the circle be \((h, k)\). Since the radius is \(13\), we can express the center in terms of the distance from the point of tangency:
\[ \sqrt{(h - 1)^2 + (k - 1)^2} = 13 \]
Squaring both sides gives:
\[ (h - 1)^2 + (k - 1)^2 = 169 \]

### Step 6: Express \(k\) in terms of \(h\)
From the slope of the radius, we have:
\[ k - 1 = -\frac{5}{12}(h - 1) \implies k = -\frac{5}{12}h + \frac{5}{12} + 1 = -\frac{5}{12}h + \frac{17}{12} \]

### Step 7: Substitute \(k\) into the circle equation
Substituting \(k\) into the distance equation:
\[ (h - 1)^2 + \left(-\frac{5}{12}h + \frac{17}{12} - 1\right)^2 = 169 \]
This simplifies to:
\[ (h - 1)^2 + \left(-\frac{5}{12}h + \frac{5}{12}\right)^2 = 169 \]
\[ (h - 1)^2 + \left(-\frac{5}{12}(h - 1)\right)^2 = 169 \]
Let \(x = h - 1\):
\[ x^2 + \left(-\frac{5}{12}x\right)^2 = 169 \]
\[ x^2 + \frac{25}{144}x^2 = 169 \]
\[ \left(1 + \frac{25}{144}\right)x^2 = 169 \]
\[ \frac{169}{144}x^2 = 169 \implies x^2 = 144 \implies x = \pm 12 \]
Thus, \(h - 1 = 12\) or \(h - 1 = -12\):
\[ h = 13 \quad \text{or} \quad h = -11 \]

### Step 8: Find corresponding \(k\) values
For \(h = 13\):
\[ k = -\frac{5}{12}(13) + \frac{17}{12} = -\frac{65}{12} + \frac{17}{12} = -\frac{48}{12} = -4 \]
For \(h = -11\):
\[ k = -\frac{5}{12}(-11) + \frac{17}{12} = \frac{55}{12} + \frac{17}{12} = \frac{72}{12} = 6 \]

### Step 9: Write the equations of the circles
The centers are \((13, -4)\) and \((-11, 6)\) with radius \(13\). The standard form of the circle equation is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Thus, the equations of the circles are:
1. \((x - 13)^2 + (y + 4)^2 = 169\)
2. \((x + 11)^2 + (y - 6)^2 = 169\)

### Final Answer
The equations of the circles are:
1. \((x - 13)^2 + (y + 4)^2 = 169\)
2. \((x + 11)^2 + (y - 6)^2 = 169\)