To find the area of the triangle formed by the tangent and normal at the point \((1, \sqrt{3})\) to the circle \(x^2 + y^2 = 4\) and the X-axis, we can follow these steps:
### Step 1: Identify the Circle's Properties
The equation of the circle is given as \(x^2 + y^2 = 4\).
- The center of the circle is at the origin \((0, 0)\).
- The radius of the circle is \(r = 2\).
### Step 2: Confirm the Point Lies on the Circle
We need to check if the point \((1, \sqrt{3})\) lies on the circle:
\[
1^2 + (\sqrt{3})^2 = 1 + 3 = 4
\]
Since this is true, the point \((1, \sqrt{3})\) is indeed on the circle.
### Step 3: Find the Slope of the Radius to the Point
The slope of the radius from the center \((0, 0)\) to the point \((1, \sqrt{3})\) is:
\[
\text{slope} = \frac{\sqrt{3} - 0}{1 - 0} = \sqrt{3}
\]
### Step 4: Find the Slope of the Tangent
The slope of the tangent line at this point is the negative reciprocal of the slope of the radius:
\[
\text{slope of tangent} = -\frac{1}{\sqrt{3}}
\]
### Step 5: Write the Equation of the Tangent Line
Using the point-slope form of the line equation, the equation of the tangent line at \((1, \sqrt{3})\) is:
\[
y - \sqrt{3} = -\frac{1}{\sqrt{3}}(x - 1)
\]
Rearranging gives:
\[
y = -\frac{1}{\sqrt{3}}x + \left(\sqrt{3} + \frac{1}{\sqrt{3}}\right) = -\frac{1}{\sqrt{3}}x + \frac{4}{\sqrt{3}}
\]
### Step 6: Find the Equation of the Normal Line
The slope of the normal line is the same as the slope of the radius, which is \(\sqrt{3}\). Thus, the equation of the normal line is:
\[
y - \sqrt{3} = \sqrt{3}(x - 1)
\]
Rearranging gives:
\[
y = \sqrt{3}x - \sqrt{3} + \sqrt{3} = \sqrt{3}x
\]
### Step 7: Find Intersection Points with the X-axis
To find the intersection of the tangent line with the X-axis, set \(y = 0\):
\[
0 = -\frac{1}{\sqrt{3}}x + \frac{4}{\sqrt{3}} \implies x = 4
\]
Thus, the point of intersection is \(C(4, 0)\).
To find the intersection of the normal line with the X-axis, set \(y = 0\):
\[
0 = \sqrt{3}x \implies x = 0
\]
Thus, the point of intersection is \(B(0, 0)\).
### Step 8: Calculate the Area of Triangle ABC
The vertices of the triangle are:
- \(A(1, \sqrt{3})\)
- \(B(0, 0)\)
- \(C(4, 0)\)
Using the formula for the area of a triangle given by vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substituting the coordinates:
\[
\text{Area} = \frac{1}{2} \left| 1(0 - 0) + 0(0 - \sqrt{3}) + 4(\sqrt{3} - 0) \right|
\]
\[
= \frac{1}{2} \left| 0 + 0 + 4\sqrt{3} \right| = \frac{1}{2} \cdot 4\sqrt{3} = 2\sqrt{3}
\]
### Conclusion
The area of the triangle formed by the tangent, normal, and the X-axis is \(2\sqrt{3}\) square units.
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