Find the equation of the circles which touch `2x-3y+1=0` at (1,1) and having radius `sqrt(13)`.

Answer

To find the equations of the circles that touch the line \(2x - 3y + 1 = 0\) at the point \((1, 1)\) and have a radius of \(\sqrt{13}\), we can follow these steps:

### Step 1: Understand the Circle's Properties
The circle must touch the line at the point \((1, 1)\) and have a radius of \(\sqrt{13}\). The center of the circle will be at a distance of \(\sqrt{13}\) from the point \((1, 1)\).

### Step 2: Find the Slope of the Given Line
The line \(2x - 3y + 1 = 0\) can be rewritten in slope-intercept form:
\[ 3y = 2x + 1 \implies y = \frac{2}{3}x + \frac{1}{3} \]
The slope of the line is \(\frac{2}{3}\). The slope of the radius at the point of tangency will be perpendicular to this line, so the slope of the radius is the negative reciprocal:
\[ \text{slope of radius} = -\frac{3}{2} \]

### Step 3: Find the Center of the Circle
Let the center of the circle be \((h, k)\). Since the center lies on the line that is perpendicular to the given line and passes through \((1, 1)\), we can write the equation of the line:
\[ k - 1 = -\frac{3}{2}(h - 1) \]
This simplifies to:
\[ k - 1 = -\frac{3}{2}h + \frac{3}{2} \implies k = -\frac{3}{2}h + \frac{5}{2} \]

### Step 4: Use the Radius to Form an Equation
The distance from the center \((h, k)\) to the point \((1, 1)\) must equal the radius \(\sqrt{13}\):
\[ \sqrt{(h - 1)^2 + (k - 1)^2} = \sqrt{13} \]
Squaring both sides gives:
\[ (h - 1)^2 + (k - 1)^2 = 13 \]

### Step 5: Substitute \(k\) into the Distance Equation
Substituting \(k = -\frac{3}{2}h + \frac{5}{2}\) into the distance equation:
\[ (h - 1)^2 + \left(-\frac{3}{2}h + \frac{5}{2} - 1\right)^2 = 13 \]
This simplifies to:
\[ (h - 1)^2 + \left(-\frac{3}{2}h + \frac{3}{2}\right)^2 = 13 \]
\[ (h - 1)^2 + \left(-\frac{3}{2}(h - 1)\right)^2 = 13 \]
\[ (h - 1)^2 + \frac{9}{4}(h - 1)^2 = 13 \]
\[ \left(1 + \frac{9}{4}\right)(h - 1)^2 = 13 \]
\[ \frac{13}{4}(h - 1)^2 = 13 \]
\[ (h - 1)^2 = 4 \implies h - 1 = \pm 2 \implies h = 3 \text{ or } h = -1 \]

### Step 6: Find Corresponding \(k\) Values
1. If \(h = 3\):
\[ k = -\frac{3}{2}(3) + \frac{5}{2} = -\frac{9}{2} + \frac{5}{2} = -2 \]
Center is \((3, -2)\).

2. If \(h = -1\):
\[ k = -\frac{3}{2}(-1) + \frac{5}{2} = \frac{3}{2} + \frac{5}{2} = 4 \]
Center is \((-1, 4)\).

### Step 7: Write the Equations of the Circles
Using the center-radius form of the circle equation \((x - h)^2 + (y - k)^2 = r^2\):
1. For center \((3, -2)\):
\[ (x - 3)^2 + (y + 2)^2 = 13 \]

2. For center \((-1, 4)\):
\[ (x + 1)^2 + (y - 4)^2 = 13 \]

### Final Answer
The equations of the circles are:
1. \((x - 3)^2 + (y + 2)^2 = 13\)
2. \((x + 1)^2 + (y - 4)^2 = 13\)