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 at \( A(-2, 3) \), \( B(2, -1) \), and \( C(4, 0) \), we can follow these steps:

### Step 1: Define the points
Let:
- \( A(-2, 3) \)
- \( B(2, -1) \)
- \( C(4, 0) \)

### Step 2: Set up the equations for distances
The circumcenter \( P(x, y) \) is equidistant from all three vertices, which means:
- Distance \( AP = BP = CP = R \) (circumradius)

The distances can be expressed as:
1. \( AP = \sqrt{(x + 2)^2 + (y - 3)^2} \)
2. \( BP = \sqrt{(x - 2)^2 + (y + 1)^2} \)
3. \( CP = \sqrt{(x - 4)^2 + y^2} \)

### Step 3: Set up equations from distances
Since \( AP = BP \), we can square both sides to eliminate the square roots:
\[ (x + 2)^2 + (y - 3)^2 = (x - 2)^2 + (y + 1)^2 \]

Expanding both sides:
- Left side:
\[ (x^2 + 4x + 4) + (y^2 - 6y + 9) = x^2 + y^2 + 4x - 6y + 13 \]
- Right side:
\[ (x^2 - 4x + 4) + (y^2 + 2y + 1) = x^2 + y^2 - 4x + 2y + 5 \]

Setting them equal:
\[ x^2 + y^2 + 4x - 6y + 13 = x^2 + y^2 - 4x + 2y + 5 \]

### Step 4: Simplify the equation
Cancel \( x^2 \) and \( y^2 \) from both sides:
\[ 4x - 6y + 13 = -4x + 2y + 5 \]
Rearranging gives:
\[ 8x - 8y + 8 = 0 \quad \Rightarrow \quad x - y + 1 = 0 \quad \Rightarrow \quad x = y - 1 \quad \text{(Equation 1)} \]

### Step 5: Set up the second equation
Now, set \( BP = CP \):
\[ (x - 2)^2 + (y + 1)^2 = (x - 4)^2 + y^2 \]

Expanding both sides:
- Left side:
\[ (x^2 - 4x + 4) + (y^2 + 2y + 1) = x^2 + y^2 - 4x + 2y + 5 \]
- Right side:
\[ (x^2 - 8x + 16) + y^2 = x^2 + y^2 - 8x + 16 \]

Setting them equal:
\[ x^2 + y^2 - 4x + 2y + 5 = x^2 + y^2 - 8x + 16 \]

### Step 6: Simplify the second equation
Cancel \( x^2 \) and \( y^2 \):
\[ -4x + 2y + 5 = -8x + 16 \]
Rearranging gives:
\[ 4x + 2y - 11 = 0 \quad \Rightarrow \quad 2y = -4x + 11 \quad \Rightarrow \quad y = -2x + \frac{11}{2} \quad \text{(Equation 2)} \]

### Step 7: Solve the equations simultaneously
Substituting Equation 1 into Equation 2:
\[ y = -2(y - 1) + \frac{11}{2} \]
Expanding:
\[ y = -2y + 2 + \frac{11}{2} \]
Combining like terms:
\[ 3y = \frac{15}{2} \quad \Rightarrow \quad y = \frac{5}{2} \]

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

### Step 8: Find the circumcenter
Thus, the circumcenter \( P \) is:
\[ P\left(\frac{3}{2}, \frac{5}{2}\right) \]

### Step 9: Calculate the circumradius
Using \( CP \) to find the circumradius \( R \):
\[ R = CP = \sqrt{\left(\frac{3}{2} - 4\right)^2 + \left(\frac{5}{2} - 0\right)^2} \]
Calculating:
\[ = \sqrt{\left(-\frac{5}{2}\right)^2 + \left(\frac{5}{2}\right)^2} = \sqrt{\frac{25}{4} + \frac{25}{4}} = \sqrt{\frac{50}{4}} = \frac{5\sqrt{2}}{2} \]

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

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