Find the equation of circle with centre
`(2, 3)` and touching the line `3x - 4y + 1 = 0`

Answer

To find the equation of the circle with center at (2, 3) that touches the line given by the equation \(3x - 4y + 1 = 0\), we can follow these steps:

### Step 1: Identify the center of the circle and the line equation
The center of the circle is given as \( (2, 3) \) and the line equation is \( 3x - 4y + 1 = 0 \).

### Step 2: Calculate the distance from the center to the line
To find the radius of the circle, we need to calculate the distance from the center of the circle to the line. The formula for the distance \( d \) from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
Here, \( A = 3 \), \( B = -4 \), \( C = 1 \), and the point \( (x_1, y_1) = (2, 3) \).

### Step 3: Substitute the values into the distance formula
Substituting the values into the formula:
\[ d = \frac{|3(2) - 4(3) + 1|}{\sqrt{3^2 + (-4)^2}} \]
Calculating the numerator:
\[ = |6 - 12 + 1| = |-5| = 5 \]
Calculating the denominator:
\[ = \sqrt{9 + 16} = \sqrt{25} = 5 \]
Thus, the distance \( d \) is:
\[ d = \frac{5}{5} = 1 \]

### Step 4: Determine the radius of the circle
The radius \( R \) of the circle is equal to the distance we just calculated:
\[ R = 1 \]

### Step 5: Write the equation of the circle
The standard form of the equation of a circle with center \( (h, k) \) and radius \( r \) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \( h = 2 \), \( k = 3 \), and \( r = 1 \):
\[ (x - 2)^2 + (y - 3)^2 = 1^2 \]
This simplifies to:
\[ (x - 2)^2 + (y - 3)^2 = 1 \]

### Step 6: Expand the equation
Expanding the equation:
\[ (x - 2)^2 + (y - 3)^2 = 1 \]
\[ x^2 - 4x + 4 + y^2 - 6y + 9 = 1 \]
Combining like terms:
\[ x^2 + y^2 - 4x - 6y + 13 = 1 \]
Rearranging gives:
\[ x^2 + y^2 - 4x - 6y + 12 = 0 \]

### Final Answer
The equation of the circle is:
\[ x^2 + y^2 - 4x - 6y + 12 = 0 \]

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