Find the co-ordinates of the circumcentre of the triangle whose vertices are :
`(i) (-2,2)`, `(2,-1)` and `(4,0)`
`(ii) (1,2)` , `(3,-4)` and `(5,-6)`

Answer

To find the coordinates of the circumcenter of the triangles given in the question, we will follow these steps for both parts of the question.

### Part (i): Vertices A(-2, 2), B(2, -1), C(4, 0)

1. **Set Up the Equations**:
We denote the circumcenter as \( O(H, K) \). The distances from \( O \) to each vertex must be equal:
\[ OA = OB = OC \]
This gives us the following equations:
\[ \sqrt{(H + 2)^2 + (K - 2)^2} = \sqrt{(H - 2)^2 + (K + 1)^2} = \sqrt{(H - 4)^2 + K^2} \]

2. **Square the Distances**:
Squaring the equations to eliminate the square roots:
\[ (H + 2)^2 + (K - 2)^2 = (H - 2)^2 + (K + 1)^2 \]
Expanding both sides:
\[ H^2 + 4H + 4 + K^2 - 4K + 4 = H^2 - 4H + 4 + K^2 + 2K + 1 \]
Simplifying gives:
\[ 8H - 6K = -3 \quad \text{(Equation 1)} \]

3. **Set Up the Second Equation**:
Now, equate the first and third distances:
\[ (H - 2)^2 + (K + 1)^2 = (H - 4)^2 + K^2 \]
Expanding gives:
\[ H^2 - 4H + 4 + K^2 + 2K + 1 = H^2 - 8H + 16 + K^2 \]
Simplifying gives:
\[ 4H + 2K = 11 \quad \text{(Equation 2)} \]

4. **Solve the System of Equations**:
We have the system:
\[ 8H - 6K = -3 \quad \text{(1)} \]
\[ 4H + 2K = 11 \quad \text{(2)} \]
From Equation (2), multiply by 3:
\[ 12H + 6K = 33 \]
Adding this to Equation (1):
\[ 20H = 30 \implies H = \frac{3}{2} \]
Substitute \( H \) back into Equation (2):
\[ 4 \left(\frac{3}{2}\right) + 2K = 11 \implies 6 + 2K = 11 \implies 2K = 5 \implies K = \frac{5}{2} \]

5. **Circumcenter Coordinates**:
The circumcenter is:
\[ O\left(\frac{3}{2}, \frac{5}{2}\right) \]

### Part (ii): Vertices A(1, 2), B(3, -4), C(5, -6)

1. **Set Up the Equations**:
Denote the circumcenter as \( O(H, K) \). The distances from \( O \) to each vertex must be equal:
\[ OA = OB = OC \]
This gives us the following equations:
\[ \sqrt{(H - 1)^2 + (K - 2)^2} = \sqrt{(H - 3)^2 + (K + 4)^2} = \sqrt{(H - 5)^2 + (K + 6)^2} \]

2. **Square the Distances**:
Squaring the equations:
\[ (H - 1)^2 + (K - 2)^2 = (H - 3)^2 + (K + 4)^2 \]
Expanding gives:
\[ H^2 - 2H + 1 + K^2 - 4K + 4 = H^2 - 6H + 9 + K^2 + 8K + 16 \]
Simplifying gives:
\[ 4H - 12K = 20 \quad \text{(Equation 3)} \]

3. **Set Up the Second Equation**:
Now, equate the first and third distances:
\[ (H - 3)^2 + (K + 4)^2 = (H - 5)^2 + (K + 6)^2 \]
Expanding gives:
\[ H^2 - 6H + 9 + K^2 + 8K + 16 = H^2 - 10H + 25 + K^2 + 12K + 36 \]
Simplifying gives:
\[ 4H - 4K = 9 \quad \text{(Equation 4)} \]

4. **Solve the System of Equations**:
We have the system:
\[ 4H - 12K = 20 \quad \text{(3)} \]
\[ 4H - 4K = 9 \quad \text{(4)} \]
Subtract Equation (4) from Equation (3):
\[ -8K = 11 \implies K = -\frac{11}{8} \]
Substitute \( K \) back into Equation (4):
\[ 4H - 4\left(-\frac{11}{8}\right) = 9 \implies 4H + \frac{44}{8} = 9 \implies 4H = 9 - \frac{11}{2} \implies 4H = \frac{18 - 11}{2} = \frac{7}{2} \]
Thus,
\[ H = \frac{7}{8} \]

5. **Circumcenter Coordinates**:
The circumcenter is:
\[ O\left(\frac{7}{8}, -\frac{11}{8}\right) \]

### Final Answers:
- For part (i): Circumcenter is \( O\left(\frac{3}{2}, \frac{5}{2}\right) \)
- For part (ii): Circumcenter is \( O\left(\frac{7}{8}, -\frac{11}{8}\right) \)