Using integration, find the area of the triangle formed by negative x-axis and tangent and normal to the circle `x^2 + y^2 = 9` at `(-1,2sqrt2)`.

Options

A. 6
B. `9 sqrt(2)`
C. `5 sqrt(2)`
D. 12

Answer

To find the area of the triangle formed by the negative x-axis and the tangent and normal to the circle \(x^2 + y^2 = 9\) at the point \((-1, 2\sqrt{2})\), we will follow these steps:

### Step 1: Find the Slope of the Normal
The center of the circle is at the origin \((0, 0)\) and the point of tangency is \((-1, 2\sqrt{2})\). The slope of the line connecting the center to the point is given by:
\[ \text{slope of OP} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2\sqrt{2} - 0}{-1 - 0} = \frac{2\sqrt{2}}{-1} = -2\sqrt{2} \]
The slope of the normal line is the same as the slope of OP, which is \(-2\sqrt{2}\).

### Step 2: Find the Slope of the Tangent
The slope of the tangent line is the negative reciprocal of the slope of the normal line:
\[ \text{slope of tangent} = \frac{1}{-2\sqrt{2}} = \frac{1}{2\sqrt{2}} \]

### Step 3: Write the Equation of the Tangent Line
Using the point-slope form of the line equation \(y - y_1 = m(x - x_1)\):
\[ y - 2\sqrt{2} = \frac{1}{2\sqrt{2}}(x + 1) \]
Rearranging gives:
\[ y = \frac{1}{2\sqrt{2}}x + 2\sqrt{2} - \frac{1}{2\sqrt{2}} = \frac{1}{2\sqrt{2}}x + \frac{4\sqrt{2}}{2} - \frac{1}{2\sqrt{2}} = \frac{1}{2\sqrt{2}}x + \frac{3\sqrt{2}}{2} \]

### Step 4: Write the Equation of the Normal Line
Using the point-slope form again:
\[ y - 2\sqrt{2} = -2\sqrt{2}(x + 1) \]
Rearranging gives:
\[ y = -2\sqrt{2}x - 2\sqrt{2} + 2\sqrt{2} = -2\sqrt{2}x \]

### Step 5: Find the Points of Intersection with the Negative x-axis
1. **Tangent Line with x-axis**: Set \(y = 0\):
\[ 0 = \frac{1}{2\sqrt{2}}x + \frac{3\sqrt{2}}{2} \]
Solving for \(x\):
\[ \frac{1}{2\sqrt{2}}x = -\frac{3\sqrt{2}}{2} \implies x = -3\sqrt{2} \]
So, the point is \((-3\sqrt{2}, 0)\).

2. **Normal Line with x-axis**: Set \(y = 0\):
\[ 0 = -2\sqrt{2}x \implies x = 0 \]
So, the point is \((0, 0)\).

### Step 6: Area of the Triangle
The vertices of the triangle are \((-3\sqrt{2}, 0)\), \((0, 0)\), and \((-1, 2\sqrt{2})\). The area \(A\) of the triangle can be calculated using the formula:
\[ A = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right| \]
Substituting the points:
\[ A = \frac{1}{2} \left| (-3\sqrt{2})(0 - 2\sqrt{2}) + (0)(2\sqrt{2} - 0) + (-1)(0 - 0) \right| \]
\[ = \frac{1}{2} \left| (-3\sqrt{2})(-2\sqrt{2}) \right| = \frac{1}{2} \left| 6 \cdot 2 \right| = \frac{1}{2} \cdot 12 = 6 \]

### Final Area Calculation
Thus, the area of the triangle is:
\[ \text{Area} = 6 \text{ square units} \]