The angle at which the circles `x^(2)+y^(2)+8x-2y-9=0` and `x^(2)+y^(2) -2x +8y-7=0` intersect is ………..

Answer

To find the angle at which the circles \( x^2 + y^2 + 8x - 2y - 9 = 0 \) and \( x^2 + y^2 - 2x + 8y - 7 = 0 \) intersect, we will follow these steps:

### Step 1: Rewrite the equations of the circles in standard form
1. For the first circle:
\[ x^2 + y^2 + 8x - 2y - 9 = 0 \]
We can complete the square:
\[ (x^2 + 8x) + (y^2 - 2y) = 9 \]
Completing the square for \(x\):
\[ x^2 + 8x = (x + 4)^2 - 16 \]
Completing the square for \(y\):
\[ y^2 - 2y = (y - 1)^2 - 1 \]
Thus, we have:
\[ (x + 4)^2 - 16 + (y - 1)^2 - 1 = 9 \]
Simplifying gives:
\[ (x + 4)^2 + (y - 1)^2 = 26 \]
So, the center \(C_1\) is \((-4, 1)\) and the radius \(r_1 = \sqrt{26}\).

2. For the second circle:
\[ x^2 + y^2 - 2x + 8y - 7 = 0 \]
Completing the square:
\[ (x^2 - 2x) + (y^2 + 8y) = 7 \]
Completing the square for \(x\):
\[ x^2 - 2x = (x - 1)^2 - 1 \]
Completing the square for \(y\):
\[ y^2 + 8y = (y + 4)^2 - 16 \]
Thus, we have:
\[ (x - 1)^2 - 1 + (y + 4)^2 - 16 = 7 \]
Simplifying gives:
\[ (x - 1)^2 + (y + 4)^2 = 24 \]
So, the center \(C_2\) is \((1, -4)\) and the radius \(r_2 = \sqrt{24}\).

### Step 2: Find the distance between the centers \(C_1\) and \(C_2\)
The distance \(d\) between the centers \(C_1(-4, 1)\) and \(C_2(1, -4)\) is given by:
\[ d = \sqrt{(1 - (-4))^2 + (-4 - 1)^2} = \sqrt{(1 + 4)^2 + (-5)^2} = \sqrt{5^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \]

### Step 3: Use the formula for the angle between two circles
The angle \(\theta\) between two circles can be calculated using the formula:
\[ \cos \theta = \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2} \]
Substituting the values we found:
- \(r_1^2 = 26\)
- \(r_2^2 = 24\)
- \(d^2 = 50\)

Thus,
\[ \cos \theta = \frac{26 + 24 - 50}{2 \cdot \sqrt{26} \cdot \sqrt{24}} = \frac{0}{2 \cdot \sqrt{26} \cdot \sqrt{24}} = 0 \]

### Step 4: Determine the angle
Since \(\cos \theta = 0\), it follows that:
\[ \theta = 90^\circ \]

### Final Answer
The angle at which the circles intersect is \(90^\circ\).

---