Points (1,0) and (2,0) are taken on the axis of x. On the line joining these two points, an equilateral triangle is described, its vertex being in the positive quadrant . Find the equations of circles described on the side of the triangle as diameter.

Answer

To solve the problem step by step, we will follow these instructions:

### 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 Third Vertex of the Equilateral Triangle
Let the third vertex of the equilateral triangle be denoted as \( C(H, K) \). Since the triangle is equilateral, the distance from \( A \) to \( C \) must equal the distance from \( B \) to \( C \).

Using the distance formula:
\[ AC = BC \]
This translates to:
\[ \sqrt{(H - 1)^2 + (K - 0)^2} = \sqrt{(H - 2)^2 + (K - 0)^2} \]
Squaring both sides, we get:
\[ (H - 1)^2 + K^2 = (H - 2)^2 + K^2 \]
Cancelling \( K^2 \) from both sides, we simplify to:
\[ (H - 1)^2 = (H - 2)^2 \]

### Step 3: Expand and Simplify
Expanding both sides:
\[ H^2 - 2H + 1 = H^2 - 4H + 4 \]
Cancelling \( H^2 \) from both sides:
\[ -2H + 1 = -4H + 4 \]
Rearranging gives:
\[ 2H = 3 \quad \Rightarrow \quad H = \frac{3}{2} \]

### Step 4: Find K
Now substitute \( H = \frac{3}{2} \) back into one of the distance equations to find \( K \). Using the equation:
\[ (H - 1)^2 + K^2 = 1^2 \]
Substituting \( H \):
\[ \left(\frac{3}{2} - 1\right)^2 + K^2 = 1 \]
This simplifies to:
\[ \left(\frac{1}{2}\right)^2 + K^2 = 1 \quad \Rightarrow \quad \frac{1}{4} + K^2 = 1 \]
Thus:
\[ K^2 = 1 - \frac{1}{4} = \frac{3}{4} \quad \Rightarrow \quad K = \frac{\sqrt{3}}{2} \]

### Step 5: Identify the Coordinates of Point C
The coordinates of point \( C \) are:
\[ C\left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right) \]

### Step 6: Find the Equations of the Circles
We need to find the equations of circles with diameters formed by the sides of the triangle.

1. **Circle with diameter AB (1,0) and (2,0)**:
Using the diameter form of the circle:
\[ (x - 1)(x - 2) + (y - 0)(y - 0) = 0 \]
Expanding gives:
\[ x^2 - 3x + 2 + y^2 = 0 \quad \Rightarrow \quad x^2 + y^2 - 3x + 2 = 0 \]

2. **Circle with diameter AC (1,0) and (3/2, sqrt(3)/2)**:
Using the diameter form:
\[ (x - 1)\left(x - \frac{3}{2}\right) + (y - 0)\left(y - \frac{\sqrt{3}}{2}\right) = 0 \]
Expanding gives:
\[ x^2 - \frac{5}{2}x + \frac{3}{2} + y^2 - \frac{\sqrt{3}}{2}y = 0 \]

3. **Circle with diameter BC (2,0) and (3/2, sqrt(3)/2)**:
Using the diameter form:
\[ (x - 2)\left(x - \frac{3}{2}\right) + (y - 0)\left(y - \frac{\sqrt{3}}{2}\right) = 0 \]
Expanding gives:
\[ x^2 - \frac{7}{2}x + 3 + y^2 - \frac{\sqrt{3}}{2}y = 0 \]

### Final Equations of the Circles
1. Circle with diameter AB:
\[ x^2 + y^2 - 3x + 2 = 0 \]
2. Circle with diameter AC:
\[ x^2 + y^2 - \frac{5}{2}x - \frac{\sqrt{3}}{2}y + \frac{3}{2} = 0 \]
3. Circle with diameter BC:
\[ x^2 + y^2 - \frac{7}{2}x - \frac{\sqrt{3}}{2}y + 3 = 0 \]