Find the circumcentre and circumradius of the triangle whose vertices are `(-2, 3), (2, -1)` and (4, 0).

Answer

To find the circumcenter and circumradius of the triangle with vertices A(-2, 3), B(2, -1), and C(4, 0), we will follow these steps:

### Step 1: Set Up the Problem
We denote the circumcenter as O(x, y). The circumcenter is equidistant from all three vertices of the triangle.

### Step 2: Write the Distance Equations
From the circumcenter O to each vertex, we can write the following equations based on the distance formula:

1. From O to A:
\[ OA^2 = (x + 2)^2 + (y - 3)^2 \]

2. From O to B:
\[ OB^2 = (x - 2)^2 + (y + 1)^2 \]

3. From O to C:
\[ OC^2 = (x - 4)^2 + (y - 0)^2 \]

Since OA = OB = OC = R (the circumradius), we can set up the following equations:

### Step 3: Equate OA^2 and OB^2
Setting OA^2 = OB^2 gives:
\[ (x + 2)^2 + (y - 3)^2 = (x - 2)^2 + (y + 1)^2 \]

Expanding both sides:
\[ x^2 + 4x + 4 + y^2 - 6y + 9 = x^2 - 4x + 4 + y^2 + 2y + 1 \]

Canceling out \(x^2\) and \(y^2\):
\[ 4x + 4 - 6y + 9 = -4x + 4 + 2y + 1 \]

Rearranging the equation:
\[ 8x - 8y + 8 = 0 \quad \Rightarrow \quad x - y + 1 = 0 \quad \Rightarrow \quad y = x + 1 \quad \text{(Equation 1)} \]

### Step 4: Equate OB^2 and OC^2
Now set OB^2 = OC^2:
\[ (x - 2)^2 + (y + 1)^2 = (x - 4)^2 + y^2 \]

Expanding both sides:
\[ x^2 - 4x + 4 + y^2 + 2y + 1 = x^2 - 8x + 16 + y^2 \]

Canceling out \(x^2\) and \(y^2\):
\[ -4x + 5 + 2y = -8x + 16 \]

Rearranging the equation:
\[ 4x + 2y - 11 = 0 \quad \Rightarrow \quad 2y = -4x + 11 \quad \Rightarrow \quad y = -2x + \frac{11}{2} \quad \text{(Equation 2)} \]

### Step 5: Solve the System of Equations
Now we have two equations:
1. \(y = x + 1\)
2. \(y = -2x + \frac{11}{2}\)

Setting them equal to each other:
\[ x + 1 = -2x + \frac{11}{2} \]

Solving for x:
\[ 3x = \frac{11}{2} - 1 \]
\[ 3x = \frac{11}{2} - \frac{2}{2} = \frac{9}{2} \]
\[ x = \frac{3}{2} \]

Substituting \(x\) back into Equation 1 to find \(y\):
\[ y = \frac{3}{2} + 1 = \frac{5}{2} \]

Thus, the circumcenter O is:
\[ O\left(\frac{3}{2}, \frac{5}{2}\right) \]

### Step 6: Calculate the Circumradius
We can find the circumradius R using the distance from O to any vertex, say C(4, 0):
\[ R^2 = OC^2 = \left(\frac{3}{2} - 4\right)^2 + \left(\frac{5}{2} - 0\right)^2 \]
\[ = \left(-\frac{5}{2}\right)^2 + \left(\frac{5}{2}\right)^2 = \frac{25}{4} + \frac{25}{4} = \frac{50}{4} = \frac{25}{2} \]
\[ R = \sqrt{\frac{25}{2}} = \frac{5\sqrt{2}}{2} \]

### Final Answer
The circumcenter is \(O\left(\frac{3}{2}, \frac{5}{2}\right)\) and the circumradius is \(R = \frac{5\sqrt{2}}{2}\).

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