Find the equation of the circle which touches both the axes and the line `3x-4y+8=0` and lies in the third quadrant.

Options

A. `x^(2)+y^(2)-4x-4y+4=0`
B. `x^(2)+y^(2)+4x+4y+4=0`
C. `x^(2)+y^(2)+2x+2y+4=0`
D. none of these

Explanation

To find the equation of the circle that touches both axes and the line \(3x - 4y + 8 = 0\) and lies in the third quadrant, we can follow these steps:

### Step 1: Understand the Circle's Properties
Since the circle touches both axes and lies in the third quadrant, its center will be at \((-a, -a)\) where \(a\) is the radius of the circle.

### Step 2: 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 \(h = -a\), \(k = -a\), and \(r = a\), we get:
\[ (x + a)^2 + (y + a)^2 = a^2 \]

### Step 3: Expand the Equation
Expanding the equation:
\[ (x + a)^2 + (y + a)^2 = a^2 \]
\[ x^2 + 2ax + a^2 + y^2 + 2ay + a^2 = a^2 \]
Combining like terms:
\[ x^2 + y^2 + 2ax + 2ay + 2a^2 = a^2 \]
Rearranging gives:
\[ x^2 + y^2 + 2ax + 2ay + a^2 = 0 \]

### Step 4: Find the Distance from the Center to the Line
The distance \(d\) from the center of the circle \((-a, -a)\) to the line \(3x - 4y + 8 = 0\) can be calculated using the formula:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
where \(A = 3\), \(B = -4\), \(C = 8\), and \((x_1, y_1) = (-a, -a)\).

Substituting these values:
\[ d = \frac{|3(-a) - 4(-a) + 8|}{\sqrt{3^2 + (-4)^2}} = \frac{|-3a + 4a + 8|}{\sqrt{9 + 16}} = \frac{|a + 8|}{5} \]
Since \(a\) is positive, we have:
\[ d = \frac{a + 8}{5} \]

### Step 5: Set the Distance Equal to the Radius
Since the circle touches the line, the distance \(d\) must equal the radius \(a\):
\[ \frac{a + 8}{5} = a \]
Multiplying both sides by 5:
\[ a + 8 = 5a \]
Rearranging gives:
\[ 8 = 5a - a \implies 8 = 4a \implies a = 2 \]

### Step 6: Substitute \(a\) Back into the Circle's Equation
Now substituting \(a = 2\) back into the circle's equation:
\[ x^2 + y^2 + 2(2)x + 2(2)y + 2^2 = 0 \]
This simplifies to:
\[ x^2 + y^2 + 4x + 4y + 4 = 0 \]

### Final Equation of the Circle
Thus, the equation of the circle is:
\[ x^2 + y^2 + 4x + 4y + 4 = 0 \]