Find the equation of the circle concentric with the circle `x^(2)+y^(2)-8x+6y-5=0` and passing through the point (-2,-7),

Answer

To find the equation of the circle that is concentric with the circle given by the equation \(x^2 + y^2 - 8x + 6y - 5 = 0\) and passes through the point \((-2, -7)\), we can follow these steps:

### Step 1: Identify the center of the given circle
The general form of a circle's equation is:
\[ x^2 + y^2 + 2ax + 2by + c = 0 \]
From the given equation, we can rewrite it as:
\[ x^2 + y^2 - 8x + 6y - 5 = 0 \]
Here, \(2a = -8\) and \(2b = 6\). Therefore:
\[ a = -4 \quad \text{and} \quad b = 3 \]
The center of the circle is given by the coordinates \((-a, -b)\):
\[ \text{Center} = (4, -3) \]

### Step 2: Write the equation of the concentric circle
Since the new circle is concentric with the given circle, it will have the same center \((4, -3)\). The general equation of a circle with center \((h, k)\) and radius \(r\) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting the center:
\[ (x - 4)^2 + (y + 3)^2 = r^2 \]

### Step 3: Find the radius using the point \((-2, -7)\)
The circle must pass through the point \((-2, -7)\). We will substitute this point into the equation to find \(r^2\):
\[ (-2 - 4)^2 + (-7 + 3)^2 = r^2 \]
Calculating each term:
\[ (-6)^2 + (-4)^2 = r^2 \]
\[ 36 + 16 = r^2 \]
\[ r^2 = 52 \]

### Step 4: Write the final equation of the circle
Now we can substitute \(r^2\) back into the equation of the circle:
\[ (x - 4)^2 + (y + 3)^2 = 52 \]

### Step 5: Expand the equation
Expanding the equation:
\[ (x - 4)^2 + (y + 3)^2 = 52 \]
\[ x^2 - 8x + 16 + y^2 + 6y + 9 = 52 \]
Combining like terms:
\[ x^2 + y^2 - 8x + 6y + 25 - 52 = 0 \]
\[ x^2 + y^2 - 8x + 6y - 27 = 0 \]

### Final Answer
The equation of the circle is:
\[ x^2 + y^2 - 8x + 6y - 27 = 0 \]