Find the equation of the circle, which passes through the point (5,4) and is concentric with the circle `x^(2)+y^(2)-8x -1 2 y + 15 = 0`.

Answer

To find the equation of the circle that passes through the point (5, 4) and is concentric with the given circle \(x^2 + y^2 - 8x - 12y + 15 = 0\), we can follow these steps:

### Step 1: Identify the center and radius of the given circle
The equation of the given circle can be rearranged to find its center and radius.

Starting with:
\[ x^2 + y^2 - 8x - 12y + 15 = 0 \]

### Step 2: Complete the square for \(x\) and \(y\)
Group the \(x\) terms and the \(y\) terms:
\[ (x^2 - 8x) + (y^2 - 12y) = -15 \]

Now, complete the square:
- For \(x^2 - 8x\), add and subtract \(16\) (which is \((\frac{8}{2})^2\)):
\[ x^2 - 8x = (x - 4)^2 - 16 \]
- For \(y^2 - 12y\), add and subtract \(36\) (which is \((\frac{12}{2})^2\)):
\[ y^2 - 12y = (y - 6)^2 - 36 \]

Now substitute back into the equation:
\[ ((x - 4)^2 - 16) + ((y - 6)^2 - 36) = -15 \]
This simplifies to:
\[ (x - 4)^2 + (y - 6)^2 - 52 = -15 \]
\[ (x - 4)^2 + (y - 6)^2 = 37 \]

### Step 3: Identify the center and radius
From the equation \((x - 4)^2 + (y - 6)^2 = 37\), we can see that:
- The center of the circle is \((4, 6)\)
- The radius \(r\) is \(\sqrt{37}\)

### Step 4: Write the equation of the required circle
Since the required circle is concentric with the given circle, it will have the same center \((4, 6)\) but a different radius.

Let the radius of the required circle be \(R\). Since it passes through the point \((5, 4)\), we can find \(R\) using the distance formula:
\[ R = \sqrt{(5 - 4)^2 + (4 - 6)^2} \]
Calculating this gives:
\[ R = \sqrt{(1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \]

### Step 5: Write the final equation of the required circle
The equation of the required circle is:
\[ (x - 4)^2 + (y - 6)^2 = R^2 \]
Substituting \(R^2 = 5\):
\[ (x - 4)^2 + (y - 6)^2 = 5 \]

### Final Answer
The equation of the required circle is:
\[ (x - 4)^2 + (y - 6)^2 = 5 \]