Find the angle of intersection of the circles `x^2+y^2-6x+4y+11=0andx^2+y^2-4x+6y+9=0`

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Explanation

To find the angle of intersection of the circles given by the equations \(x^2 + y^2 - 6x + 4y + 11 = 0\) and \(x^2 + y^2 - 4x + 6y + 9 = 0\), we will follow these steps:

### Step 1: Identify the center and radius of the first circle
The general form of a circle is given by:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
For the first circle, we have:
\[ x^2 + y^2 - 6x + 4y + 11 = 0 \]
From this, we can identify:
- \(2g = -6 \Rightarrow g = -3\)
- \(2f = 4 \Rightarrow f = 2\)
- \(c = 11\)

The center \((h_1, k_1)\) of the first circle is:
\[ (h_1, k_1) = (-g, -f) = (3, -2) \]

The radius \(r_1\) is given by:
\[ r_1 = \sqrt{g^2 + f^2 - c} = \sqrt{(-3)^2 + (2)^2 - 11} = \sqrt{9 + 4 - 11} = \sqrt{2} \]

### Step 2: Identify the center and radius of the second circle
For the second circle, we have:
\[ x^2 + y^2 - 4x + 6y + 9 = 0 \]
Identifying the coefficients:
- \(2g = -4 \Rightarrow g = -2\)
- \(2f = 6 \Rightarrow f = 3\)
- \(c = 9\)

The center \((h_2, k_2)\) of the second circle is:
\[ (h_2, k_2) = (-g, -f) = (2, -3) \]

The radius \(r_2\) is given by:
\[ r_2 = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + (3)^2 - 9} = \sqrt{4 + 9 - 9} = \sqrt{4} = 2 \]

### Step 3: Calculate the distance between the centers of the circles
The distance \(d\) between the centers \((h_1, k_1)\) and \((h_2, k_2)\) is calculated using the distance formula:
\[ d = \sqrt{(h_2 - h_1)^2 + (k_2 - k_1)^2} = \sqrt{(2 - 3)^2 + (-3 + 2)^2} = \sqrt{(-1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

### Step 4: Use the formula for the angle of intersection
The formula for the cosine of the angle of intersection \(\theta\) of two circles is given by:
\[ \cos \theta = \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2} \]
Substituting the values:
- \(r_1^2 = (\sqrt{2})^2 = 2\)
- \(r_2^2 = 2^2 = 4\)
- \(d^2 = (\sqrt{2})^2 = 2\)

Thus,
\[ \cos \theta = \frac{2 + 4 - 2}{2 \cdot \sqrt{2} \cdot 2} = \frac{4}{4\sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Step 5: Calculate the angle \(\theta\)
From \(\cos \theta = \frac{1}{\sqrt{2}}\), we find:
\[ \theta = 45^\circ \]

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

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