(-1,2) is the vertex of an equilateral triangle whose centroid is (1, 1), then the equation of its circumcircle is

Options

A. `x^(2)+y^(2)+2x+2y-3=0`
B. `x^(2)+y^(2)+2x-2y-3=0`
C. `x^(2)+y^(2)-2x-2y-3=0`
D. none

Answer

To find the equation of the circumcircle of the equilateral triangle with vertex at (-1, 2) and centroid at (1, 1), we will follow these steps:

### Step 1: Identify the centroid and vertex
The centroid (G) of the triangle is given as (1, 1) and one vertex (A) is given as (-1, 2).

### Step 2: Use the centroid formula
The centroid of a triangle with vertices A(x1, y1), B(x2, y2), and C(x3, y3) is given by:
\[ G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]
Here, we know G(1, 1) and A(-1, 2). Let the other two vertices be B(x2, y2) and C(x3, y3).

### Step 3: Set up equations based on the centroid
From the centroid formula, we can write:
\[ 1 = \frac{-1 + x_2 + x_3}{3} \quad \text{(1)} \]
\[ 1 = \frac{2 + y_2 + y_3}{3} \quad \text{(2)} \]

### Step 4: Solve for x2 and x3 using equation (1)
Multiplying equation (1) by 3 gives:
\[ 3 = -1 + x_2 + x_3 \]
\[ x_2 + x_3 = 4 \quad \text{(3)} \]

### Step 5: Solve for y2 and y3 using equation (2)
Multiplying equation (2) by 3 gives:
\[ 3 = 2 + y_2 + y_3 \]
\[ y_2 + y_3 = 1 \quad \text{(4)} \]

### Step 6: Calculate the circumradius
The circumradius (R) can be calculated using the distance from the centroid (1, 1) to the vertex (-1, 2):
\[ R = \sqrt{(1 - (-1))^2 + (1 - 2)^2} \]
\[ R = \sqrt{(1 + 1)^2 + (1 - 2)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \]

### Step 7: Write the equation of the circumcircle
The general equation of a circle with center (h, k) and radius R is:
\[ (x - h)^2 + (y - k)^2 = R^2 \]
Substituting h = 1, k = 1, and R = √5, we get:
\[ (x - 1)^2 + (y - 1)^2 = (\sqrt{5})^2 \]
\[ (x - 1)^2 + (y - 1)^2 = 5 \]

### Step 8: Expand the equation
Expanding the left side:
\[ (x^2 - 2x + 1) + (y^2 - 2y + 1) = 5 \]
\[ x^2 + y^2 - 2x - 2y + 2 = 5 \]
\[ x^2 + y^2 - 2x - 2y - 3 = 0 \]

### Final Answer
The equation of the circumcircle is:
\[ x^2 + y^2 - 2x - 2y - 3 = 0 \]