To solve the problem, we need to find the equations of the circles described on the sides of the equilateral triangle formed by the points (1, 0) and (2, 0) as diameters. Let's go through the solution step by step.
### Step 1: Identify the Points
We have two points on the x-axis:
- Point A: (1, 0)
- Point B: (2, 0)
### Step 2: Find the Coordinates of the Third Vertex
An equilateral triangle has all sides equal and the angles between them are 60 degrees. The length of side AB is:
\[ AB = 2 - 1 = 1 \]
To find the coordinates of the third vertex (C) of the triangle, we can use the fact that the height of an equilateral triangle can be calculated as:
\[ \text{Height} = \frac{\sqrt{3}}{2} \times \text{side length} = \frac{\sqrt{3}}{2} \times 1 = \frac{\sqrt{3}}{2} \]
The coordinates of point C can be calculated as follows:
- The midpoint of AB is:
\[ \left( \frac{1 + 2}{2}, 0 \right) = \left( \frac{3}{2}, 0 \right) \]
- The height is added to the y-coordinate of the midpoint:
\[ C = \left( \frac{3}{2}, \frac{\sqrt{3}}{2} \right) \]
### Step 3: Find the Circle with AB as Diameter
The center of the circle with diameter AB is the midpoint of AB:
\[ \text{Midpoint of AB} = \left( \frac{1 + 2}{2}, \frac{0 + 0}{2} \right) = \left( \frac{3}{2}, 0 \right) \]
The radius is half the length of AB:
\[ \text{Radius} = \frac{1}{2} \]
The equation of the circle is given by:
\[
\left( x - \frac{3}{2} \right)^2 + (y - 0)^2 = \left( \frac{1}{2} \right)^2
\]
Simplifying this, we get:
\[
\left( x - \frac{3}{2} \right)^2 + y^2 = \frac{1}{4}
\]
### Step 4: Find the Circle with AC as Diameter
Now, we find the midpoint of AC:
- A = (1, 0)
- C = \(\left( \frac{3}{2}, \frac{\sqrt{3}}{2} \right)\)
The midpoint of AC is:
\[
E = \left( \frac{1 + \frac{3}{2}}{2}, \frac{0 + \frac{\sqrt{3}}{2}}{2} \right) = \left( \frac{5}{4}, \frac{\sqrt{3}}{4} \right)
\]
The radius is again half the length of AC. The length of AC can be calculated using the distance formula:
\[
AC = \sqrt{\left( \frac{3}{2} - 1 \right)^2 + \left( \frac{\sqrt{3}}{2} - 0 \right)^2} = \sqrt{\left( \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1
\]
Thus, the radius is:
\[ \text{Radius} = \frac{1}{2} \]
The equation of the circle is:
\[
\left( x - \frac{5}{4} \right)^2 + \left( y - \frac{\sqrt{3}}{4} \right)^2 = \left( \frac{1}{2} \right)^2
\]
### Step 5: Find the Circle with BC as Diameter
Now, we find the midpoint of BC:
- B = (2, 0)
- C = \(\left( \frac{3}{2}, \frac{\sqrt{3}}{2} \right)\)
The midpoint of BC is:
\[
F = \left( \frac{2 + \frac{3}{2}}{2}, \frac{0 + \frac{\sqrt{3}}{2}}{2} \right) = \left( \frac{7}{4}, \frac{\sqrt{3}}{4} \right)
\]
The radius is again half the length of BC. The length of BC can be calculated as:
\[
BC = \sqrt{\left( 2 - \frac{3}{2} \right)^2 + \left( 0 - \frac{\sqrt{3}}{2} \right)^2} = \sqrt{\left( \frac{1}{2} \right)^2 + \left( -\frac{\sqrt{3}}{2} \right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1
\]
Thus, the radius is:
\[ \text{Radius} = \frac{1}{2} \]
The equation of the circle is:
\[
\left( x - \frac{7}{4} \right)^2 + \left( y - \frac{\sqrt{3}}{4} \right)^2 = \left( \frac{1}{2} \right)^2
\]
### Final Equations of the Circles:
1. Circle with diameter AB:
\[
\left( x - \frac{3}{2} \right)^2 + y^2 = \frac{1}{4}
\]
2. Circle with diameter AC:
\[
\left( x - \frac{5}{4} \right)^2 + \left( y - \frac{\sqrt{3}}{4} \right)^2 = \frac{1}{4}
\]
3. Circle with diameter BC:
\[
\left( x - \frac{7}{4} \right)^2 + \left( y - \frac{\sqrt{3}}{4} \right)^2 = \frac{1}{4}
\]
