Find the angle of intersection of the curves :
(ii) `x^(2)+y^(2)-4x-1=0 and x^(2)+y^(2)-2y-9=0.`

Answer

To find the angle of intersection of the curves given by the equations \(x^2 + y^2 - 4x - 1 = 0\) and \(x^2 + y^2 - 2y - 9 = 0\), we will follow these steps:

### Step 1: Rewrite the equations in standard form
The equations of the circles can be rewritten in standard form by completing the square.

1. **First Circle:**
\[ x^2 + y^2 - 4x - 1 = 0 \]
Completing the square for \(x\):
\[ (x^2 - 4x) + y^2 = 1 \implies (x - 2)^2 + y^2 = 5 \]
This represents a circle with center \(C_1(2, 0)\) and radius \(r_1 = \sqrt{5}\).

2. **Second Circle:**
\[ x^2 + y^2 - 2y - 9 = 0 \]
Completing the square for \(y\):
\[ x^2 + (y^2 - 2y) = 9 \implies x^2 + (y - 1)^2 = 10 \]
This represents a circle with center \(C_2(0, 1)\) and radius \(r_2 = \sqrt{10}\).

### Step 2: Calculate the distance between the centers
The distance \(d\) between the centers \(C_1(2, 0)\) and \(C_2(0, 1)\) can be calculated using the distance formula:
\[ d = \sqrt{(2 - 0)^2 + (0 - 1)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \]

### Step 3: Use the formula for the angle of intersection
The angle \(\theta\) between two intersecting circles can be found using the formula:
\[ \cos \theta = \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2} \]
Substituting the values:
- \(r_1 = \sqrt{5}\) so \(r_1^2 = 5\)
- \(r_2 = \sqrt{10}\) so \(r_2^2 = 10\)
- \(d^2 = 5\)

Now substituting these into the formula:
\[ \cos \theta = \frac{5 + 10 - 5}{2 \cdot \sqrt{5} \cdot \sqrt{10}} = \frac{10}{2 \cdot \sqrt{5} \cdot \sqrt{10}} = \frac{10}{2 \cdot \sqrt{50}} = \frac{10}{2 \cdot 5\sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Step 4: Find the angle \(\theta\)
From \(\cos \theta = \frac{1}{\sqrt{2}}\), we find:
\[ \theta = \frac{\pi}{4} \text{ radians} = 45^\circ \]

### Final Answer
The angle of intersection of the curves is \(45^\circ\).

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