Find the equation of the circle passing through (-2,14) and concentric with the circle
`x^(2)+y^(2)-6x-4y-12=0`

Answer

To find the equation of the circle that passes through the point (-2, 14) and is concentric with the circle given by the equation \( x^2 + y^2 - 6x - 4y - 12 = 0 \), we will follow these steps:

### Step 1: Identify the center of the given circle
The general form of the equation of a circle is:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From the given equation \( x^2 + y^2 - 6x - 4y - 12 = 0 \), we can identify:
- \( 2g = -6 \) → \( g = -3 \)
- \( 2f = -4 \) → \( f = -2 \)

Thus, the center of the given circle is:
\[ (h, k) = (-g, -f) = (3, 2) \]

### Step 2: Calculate the radius of the new circle
Since the new circle is concentric with the given circle, it will have the same center (3, 2). We need to find the radius of the new circle, which can be calculated using the distance formula from the center (3, 2) to the point (-2, 14).

The distance formula is given by:
\[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ r = \sqrt{((-2) - 3)^2 + (14 - 2)^2} \]
Calculating this:
\[ r = \sqrt{(-5)^2 + (12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \]

### Step 3: Write the equation of the new circle
The equation of a circle with center (h, k) and radius r is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \( h = 3 \), \( k = 2 \), and \( r = 13 \):
\[ (x - 3)^2 + (y - 2)^2 = 13^2 \]
This simplifies to:
\[ (x - 3)^2 + (y - 2)^2 = 169 \]

### Step 4: Expand the equation
Expanding the equation:
\[ (x^2 - 6x + 9) + (y^2 - 4y + 4) = 169 \]
Combining terms:
\[ x^2 + y^2 - 6x - 4y + 13 = 169 \]
Rearranging gives:
\[ x^2 + y^2 - 6x - 4y - 156 = 0 \]

### Final Answer
The equation of the circle is:
\[ x^2 + y^2 - 6x - 4y - 156 = 0 \]

---