The equation of the image of the circle `x^2+y^2+16x-24y+183=0` by the line mirror `4x+7y+13=0` is :

Options

A. `x^(2) +y^(2) +32x - 4y +235 = 0`
B. `x^(2) +y^(2) +32x +4y - 235 = 0`
C. `x^(2) +y^(2) +32x - 4y - 235 = 0`
D. `x^(2) +y^(2) +32x +4y +235 = 0`

Explanation

To find the equation of the image of the circle given by \(x^2 + y^2 + 16x - 24y + 183 = 0\) reflected over the line \(4x + 7y + 13 = 0\), we can follow these steps:

### Step 1: Rewrite the Circle's Equation
Start by rewriting the circle's equation in standard form. We need to complete the square for both \(x\) and \(y\).

\[ x^2 + 16x + y^2 - 24y + 183 = 0 \]

Completing the square for \(x\):
\[ x^2 + 16x = (x + 8)^2 - 64 \]

Completing the square for \(y\):
\[ y^2 - 24y = (y - 12)^2 - 144 \]

Substituting these back into the equation:
\[ (x + 8)^2 - 64 + (y - 12)^2 - 144 + 183 = 0 \]
\[ (x + 8)^2 + (y - 12)^2 - 25 = 0 \]
\[ (x + 8)^2 + (y - 12)^2 = 25 \]

### Step 2: Identify the Center and Radius
From the standard form \((x + 8)^2 + (y - 12)^2 = 5^2\), we can identify:
- Center \(C_1 = (-8, 12)\)
- Radius \(r = 5\)

### Step 3: Find the Perpendicular Line
Next, we need to find the line perpendicular to the given line \(4x + 7y + 13 = 0\) that passes through the center \(C_1\).

The slope of the line \(4x + 7y + 13 = 0\) can be found by rewriting it in slope-intercept form:
\[ 7y = -4x - 13 \implies y = -\frac{4}{7}x - \frac{13}{7} \]
Thus, the slope \(m_1 = -\frac{4}{7}\).

The slope of the perpendicular line \(m_2\) is given by:
\[ m_1 \cdot m_2 = -1 \implies m_2 = \frac{7}{4} \]

### Step 4: Equation of the Perpendicular Line
Using the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
Substituting \(C_1(-8, 12)\) and \(m_2 = \frac{7}{4}\):
\[ y - 12 = \frac{7}{4}(x + 8) \]
Multiplying through by 4 to eliminate the fraction:
\[ 4y - 48 = 7x + 56 \implies 7x - 4y + 104 = 0 \]

### Step 5: Find the Intersection Point
Now we find the intersection of the lines \(4x + 7y + 13 = 0\) and \(7x - 4y + 104 = 0\).

Multiply the first equation by 7:
\[ 28x + 49y + 91 = 0 \]
Multiply the second equation by 4:
\[ 28x - 16y + 416 = 0 \]

Now, subtract the second from the first:
\[ (49y + 16y) + (91 - 416) = 0 \implies 65y - 325 = 0 \implies y = 5 \]

Substituting \(y = 5\) back into \(4x + 7(5) + 13 = 0\):
\[ 4x + 35 + 13 = 0 \implies 4x + 48 = 0 \implies x = -12 \]

Thus, the intersection point \(C_2 = (-12, 5)\).

### Step 6: Find the Image Center
Using the midpoint formula, we know:
\[ C_1 = (-8, 12), \quad C_2 = (-12, 5) \]
Let \(h\) and \(k\) be the coordinates of the image center. The midpoint formula gives:
\[ \left(\frac{-8 + h}{2}, \frac{12 + k}{2}\right) = (-12, 5) \]

Setting up the equations:
1. \(\frac{-8 + h}{2} = -12 \implies -8 + h = -24 \implies h = -16\)
2. \(\frac{12 + k}{2} = 5 \implies 12 + k = 10 \implies k = -2\)

Thus, the image center is \(C' = (-16, -2)\).

### Step 7: Write the Equation of the Image Circle
The radius remains the same, so the equation of the image circle is:
\[ (x + 16)^2 + (y + 2)^2 = 25 \]

Expanding this:
\[ (x^2 + 32x + 256) + (y^2 + 4y + 4) = 25 \]
\[ x^2 + y^2 + 32x + 4y + 260 - 25 = 0 \]
\[ x^2 + y^2 + 32x + 4y + 235 = 0 \]

### Final Answer
The equation of the image of the circle is:
\[ \boxed{x^2 + y^2 + 32x + 4y + 235 = 0} \]