The equation of a circle which touches both the coordinate axes and the line x = 3 is

Options

A. `4x^(2) + 4y^(2) - 12 x pm 12 y + 9 = 0 `
B. `4x^(2) + 4y^(2) + 12 x pm 12 y + 9 = 0 `
C. `4x^(2) + 4y ^(2) pm 12 x - 12 y + 9 = 0 `
D. ` 4x ^(2) + 4y^(2) pm 12x + 12 y + 9 = 0 `

Answer

To find the equation of a circle that touches both the coordinate axes and the line \( x = 3 \), we can follow these steps:

### Step 1: Understand the conditions
A circle that touches both the coordinate axes means that the distance from the center of the circle to each axis is equal to the radius of the circle. Since it also touches the line \( x = 3 \), the distance from the center of the circle to this line must also equal the radius.

### Step 2: Define the center and radius
Let the center of the circle be \( (h, k) \) and the radius be \( r \). Since the circle touches the x-axis and y-axis, we have:
- The distance from the center to the x-axis is \( k \) (which equals the radius \( r \)).
- The distance from the center to the y-axis is \( h \) (which also equals the radius \( r \)).

Thus, we can write:
\[ h = r \quad \text{and} \quad k = r \]

### Step 3: Apply the condition for the line \( x = 3 \)
The distance from the center \( (h, k) \) to the line \( x = 3 \) is given by:
\[ |h - 3| = r \]
Substituting \( h = r \) into this equation gives:
\[ |r - 3| = r \]

### Step 4: Solve the absolute value equation
This absolute value equation can be split into two cases:
1. \( r - 3 = r \) (which leads to no solution)
2. \( 3 - r = r \)

From the second case:
\[ 3 = 2r \implies r = \frac{3}{2} \]

### Step 5: Find the coordinates of the center
Since \( r = \frac{3}{2} \), we can find the coordinates of the center:
\[ h = r = \frac{3}{2} \quad \text{and} \quad k = r = \frac{3}{2} \]
Thus, the center of the circle is \( \left( \frac{3}{2}, \frac{3}{2} \right) \).

### Step 6: Write the equation of the circle
The standard form of the equation of a circle is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \( h = \frac{3}{2} \), \( k = \frac{3}{2} \), and \( r = \frac{3}{2} \):
\[ \left( x - \frac{3}{2} \right)^2 + \left( y - \frac{3}{2} \right)^2 = \left( \frac{3}{2} \right)^2 \]
This simplifies to:
\[ \left( x - \frac{3}{2} \right)^2 + \left( y - \frac{3}{2} \right)^2 = \frac{9}{4} \]

### Step 7: Rearranging the equation
To express it in a more standard form, we can multiply through by 4:
\[ 4\left( x - \frac{3}{2} \right)^2 + 4\left( y - \frac{3}{2} \right)^2 = 9 \]
Expanding this gives:
\[ 4\left( x^2 - 3x + \frac{9}{4} \right) + 4\left( y^2 - 3y + \frac{9}{4} \right) = 9 \]
This simplifies to:
\[ 4x^2 + 4y^2 - 12x - 12y + 9 = 9 \]
Thus, we have:
\[ 4x^2 + 4y^2 - 12x - 12y = 0 \]

### Final Equation
The final equation of the circle is:
\[ 4x^2 + 4y^2 - 12x - 12y + 9 = 0 \]