The equation of the image of the circle `(x-3)^(2)+(y-2)=1` in the mirror x+y=19, is

Options

A. `(x+14)^(2)+(y-13)^(2)=1`
B. `(x-15)^(2)+(y-14)^(2)=1`
C. `(x-16)^(2)+(y-15)^(2)=1`
D. `(x-17)^(2)+(y-16^(2))=1`

Answer

To find the equation of the image of the circle \((x-3)^2 + (y-2)^2 = 1\) in the mirror line \(x + y = 19\), we will follow these steps:

### Step 1: Identify the center and radius of the circle
The given equation of the circle is \((x-3)^2 + (y-2)^2 = 1\).

- The center of the circle \((x_1, y_1)\) is \((3, 2)\).
- The radius \(r\) is \(\sqrt{1} = 1\).

### Step 2: Find the image of the center in the mirror
To find the image of the center \((3, 2)\) with respect to the line \(x + y = 19\), we can use the formula for the reflection of a point across a line.

The line \(x + y = 19\) can be rewritten in the standard form \(Ax + By + C = 0\) where \(A = 1\), \(B = 1\), and \(C = -19\).

Using the formula for the reflection of a point \((x_1, y_1)\) across the line \(Ax + By + C = 0\):
\[ x' = x_1 - \frac{2A(Ax_1 + By_1 + C)}{A^2 + B^2} \]
\[ y' = y_1 - \frac{2B(Ax_1 + By_1 + C)}{A^2 + B^2} \]

Substituting \(x_1 = 3\), \(y_1 = 2\), \(A = 1\), \(B = 1\), and \(C = -19\):
1. Calculate \(Ax_1 + By_1 + C\):
\[ 1 \cdot 3 + 1 \cdot 2 - 19 = 3 + 2 - 19 = -14 \]

2. Calculate \(A^2 + B^2\):
\[ 1^2 + 1^2 = 1 + 1 = 2 \]

3. Substitute into the reflection formulas:
\[ x' = 3 - \frac{2 \cdot 1 \cdot (-14)}{2} = 3 + 14 = 17 \]
\[ y' = 2 - \frac{2 \cdot 1 \cdot (-14)}{2} = 2 + 14 = 16 \]

Thus, the image of the center is \((17, 16)\).

### Step 3: Write the equation of the image circle
The radius of the image circle remains the same as the original circle, which is \(1\).

The equation of the circle with center \((17, 16)\) and radius \(1\) is given by:
\[ (x - 17)^2 + (y - 16)^2 = 1 \]

### Final Answer
The equation of the image of the circle in the mirror \(x + y = 19\) is:
\[ (x - 17)^2 + (y - 16)^2 = 1 \]

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