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`

Answer

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

### Step 1: Identify the center and radius of the circle
The general form of a circle's equation is given by:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From the given equation \(x^2 + y^2 + 16x - 24y + 183 = 0\), we can identify:
- \(2g = 16 \Rightarrow g = 8\)
- \(2f = -24 \Rightarrow f = -12\)
- \(c = 183\)

The center of the circle \((h, k)\) can be calculated as:
\[ (h, k) = (-g, -f) = (-8, 12) \]

### Step 2: Calculate the radius of the circle
The radius \(r\) is given by:
\[ r = \sqrt{g^2 + f^2 - c} \]
Calculating this:
\[ g^2 = 8^2 = 64, \quad f^2 = (-12)^2 = 144 \]
\[ r = \sqrt{64 + 144 - 183} = \sqrt{25} = 5 \]

### Step 3: Find the image of the center with respect to the line
The line mirror is given by \(4x + 7y + 13 = 0\). We will use the formula for the image of a point \((x, y)\) across the line \(ax + by + c = 0\):
\[ x' = x - \frac{2a(ax + by + c)}{a^2 + b^2}, \quad y' = y - \frac{2b(ax + by + c)}{a^2 + b^2} \]
Here, \(a = 4\), \(b = 7\), and \(c = 13\). The center of the circle is \((-8, 12)\).

Calculating \(ax + by + c\):
\[ 4(-8) + 7(12) + 13 = -32 + 84 + 13 = 65 \]

Now substituting into the formulas for \(x'\) and \(y'\):
\[ x' = -8 - \frac{2 \cdot 4 \cdot 65}{4^2 + 7^2} = -8 - \frac{520}{16 + 49} = -8 - \frac{520}{65} = -8 - 8 = -16 \]
\[ y' = 12 - \frac{2 \cdot 7 \cdot 65}{4^2 + 7^2} = 12 - \frac{910}{65} = 12 - 14 = -2 \]

Thus, the image of the center is \((-16, -2)\).

### Step 4: Write the equation of the new circle
The new circle will have the same radius \(r = 5\) and the center \((-16, -2)\). The equation of the circle is given by:
\[ (x + 16)^2 + (y + 2)^2 = 5^2 \]
Expanding this:
\[ (x^2 + 32x + 256) + (y^2 + 4y + 4) = 25 \]
Combining terms:
\[ x^2 + y^2 + 32x + 4y + 256 + 4 - 25 = 0 \]
This simplifies to:
\[ x^2 + y^2 + 32x + 4y + 235 = 0 \]

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