The equation of the circle whose center is (3,-2) and which touches the line `3x - 4y + 13 = 0` is

Options

A. A. `x^(2) + y^(2) + 6x - 4y - 23 = 0 `
B. B.` x^(2) + y^(2) - 6x - 4y - 23 = 0 `
C. C. ` x^(2) + y^(2) - 6x + 4y - 23 = 0 `
D. D. ` x^(2) + y^(2) + 6x + 4y + 23 = 0 `

Answer

To find the equation of the circle with center (3, -2) that touches the line \(3x - 4y + 13 = 0\), 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 \( (3, -2) \) and the line equation is \( 3x - 4y + 13 = 0 \).

### Step 2: Calculate the perpendicular distance from the center to the line
We will use the formula for the perpendicular distance \( d \) from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \):
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

Here, \( A = 3 \), \( B = -4 \), \( C = 13 \), \( x_1 = 3 \), and \( y_1 = -2 \).

Substituting these values into the formula:
\[ d = \frac{|3(3) - 4(-2) + 13|}{\sqrt{3^2 + (-4)^2}} \]

### Step 3: Simplify the expression
Calculating the numerator:
\[ = |9 + 8 + 13| = |30| = 30 \]

Calculating the denominator:
\[ = \sqrt{9 + 16} = \sqrt{25} = 5 \]

So, the distance \( d \) is:
\[ d = \frac{30}{5} = 6 \]

### Step 4: Determine the radius of the circle
Since the circle touches the line, the radius \( r \) of the circle is equal to the perpendicular distance calculated:
\[ r = 6 \]

### Step 5: Write the equation of the circle
The general equation of a circle with center \( (h, k) \) and radius \( r \) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]

Substituting \( h = 3 \), \( k = -2 \), and \( r = 6 \):
\[ (x - 3)^2 + (y + 2)^2 = 6^2 \]
\[ (x - 3)^2 + (y + 2)^2 = 36 \]

### Step 6: Expand the equation
Expanding the left side:
\[ (x^2 - 6x + 9) + (y^2 + 4y + 4) = 36 \]
Combining the terms:
\[ x^2 + y^2 - 6x + 4y + 13 = 36 \]
Rearranging gives:
\[ x^2 + y^2 - 6x + 4y - 23 = 0 \]

### Step 7: Identify the correct option
The final equation of the circle is:
\[ x^2 + y^2 - 6x + 4y - 23 = 0 \]

Thus, the correct option is **C**.

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