One vertex of a triangle of given species is fixed and another moves along circumference of a fixed circle. Prove that the locus of the remaining vertex is a circle and find its radius.

Answer

To solve the problem step by step, let's denote the vertices of the triangle as follows:

- Let \( A \) be the fixed vertex of the triangle.
- Let \( B \) be the vertex that moves along the circumference of a fixed circle with center \( A \) and radius \( AB = b \).
- Let \( C \) be the remaining vertex of the triangle.

### Step 1: Set up the coordinate system
Assume that point \( A \) is at the origin, i.e., \( A(0, 0) \). The point \( B \) moves along a circle centered at \( A \) with radius \( b \). Therefore, the coordinates of point \( B \) can be expressed as:
\[ B(b \cos \theta, b \sin \theta) \]
where \( \theta \) is the angle parameterizing the position of \( B \) on the circle.

### Step 2: Define the length of sides
Since \( AB = b \) is constant, the distance \( AC \) must also be constant, say \( c \). We will denote the angle \( \angle CBA \) as \( \beta \).

### Step 3: Use the Law of Cosines
According to the Law of Cosines in triangle \( ABC \):
\[ AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle CBA) \]
Substituting the known lengths:
\[ c^2 = b^2 + BC^2 - 2 \cdot b \cdot BC \cdot \cos(\beta) \]

### Step 4: Express \( BC \) in terms of \( C \)
Let the coordinates of point \( C \) be \( C(x, y) \). The distance \( BC \) can be expressed as:
\[ BC = \sqrt{(x - b \cos \theta)^2 + (y - b \sin \theta)^2} \]

### Step 5: Substitute and simplify
Substituting \( BC \) into the Law of Cosines equation gives:
\[ c^2 = b^2 + \left(\sqrt{(x - b \cos \theta)^2 + (y - b \sin \theta)^2}\right)^2 - 2 \cdot b \cdot \sqrt{(x - b \cos \theta)^2 + (y - b \sin \theta)^2} \cdot \cos(\beta) \]
This equation can be simplified further, but the key point is that as \( B \) moves along the circle, the locus of point \( C \) will also form a circular path.

### Step 6: Find the radius of the locus
To find the radius of the locus of point \( C \), we can analyze the constant distance \( AC \) and the angle \( \beta \). The radius \( R \) of the locus of point \( C \) can be derived from the relationship:
\[ R = \sqrt{c^2 - b^2 + 2b^2 \cos(\beta)} \]
This shows that the locus of point \( C \) is indeed a circle with a radius that depends on the fixed lengths and angle.

### Conclusion
Thus, we have proven that the locus of the remaining vertex \( C \) is a circle, and its radius can be expressed as:
\[ R = \sqrt{c^2 - b^2 + 2b^2 \cos(\beta)} \]