Find the angle between the circles
`x^2 + y^2 + 4x - 14y + 28 = 0` and
`x^2 + y^2 + 4x - 5 = 0`

Answer

To find the angle between the two circles given by the equations:

1. \( x^2 + y^2 + 4x - 14y + 28 = 0 \) (Circle 1)
2. \( x^2 + y^2 + 4x - 5 = 0 \) (Circle 2)

we will follow these steps:

### Step 1: Rewrite the equations in standard form
We need to convert both equations into the standard form of a circle, which is \( (x - h)^2 + (y - k)^2 = r^2 \).

**Circle 1:**
Starting with the equation:
\[ x^2 + y^2 + 4x - 14y + 28 = 0 \]
We can rearrange it:
\[ x^2 + 4x + y^2 - 14y + 28 = 0 \]
Completing the square for \(x\) and \(y\):
\[ (x^2 + 4x + 4) + (y^2 - 14y + 49) = 28 + 4 + 49 \]
This simplifies to:
\[ (x + 2)^2 + (y - 7)^2 = 25 \]
Thus, the center \(C_1\) is \((-2, 7)\) and the radius \(r_1 = 5\).

**Circle 2:**
Starting with the equation:
\[ x^2 + y^2 + 4x - 5 = 0 \]
Rearranging gives:
\[ x^2 + 4x + y^2 = 5 \]
Completing the square:
\[ (x^2 + 4x + 4) + y^2 = 5 + 4 \]
This simplifies to:
\[ (x + 2)^2 + y^2 = 9 \]
Thus, the center \(C_2\) is \((-2, 0)\) and the radius \(r_2 = 3\).

### Step 2: Calculate the distance between the centers
The distance \(d\) between the centers \(C_1\) and \(C_2\) is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ d = \sqrt{((-2) - (-2))^2 + (0 - 7)^2} = \sqrt{0 + 49} = 7 \]

### Step 3: Use the formula to find the angle between the circles
The cosine of the angle \(\theta\) between the two circles can be calculated using the formula:
\[ \cos \theta = \frac{d^2 - r_1^2 - r_2^2}{2 r_1 r_2} \]
Substituting the values:
\[ \cos \theta = \frac{7^2 - 5^2 - 3^2}{2 \cdot 5 \cdot 3} \]
Calculating:
\[ \cos \theta = \frac{49 - 25 - 9}{30} = \frac{15}{30} = \frac{1}{2} \]

### Step 4: Find the angle \(\theta\)
To find \(\theta\):
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ \]

### Final Answer
The angle between the two circles is \(60^\circ\).

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