Find the circumcentre of the triangle whose vertices are (-2, -3), (-1, 0) and (7, -6). Also find the radius of the circumircle.

Answer

To find the circumcenter of the triangle with vertices A(-2, -3), B(-1, 0), and C(7, -6), we will follow these steps:

### Step 1: Identify the vertices of the triangle
The vertices of the triangle are:
- A = (-2, -3)
- B = (-1, 0)
- C = (7, -6)

### Step 2: Set up the circumcenter equations
The circumcenter is the point (x, y) that is equidistant from all three vertices. We will use the distance formula to set up equations based on the distances from the circumcenter to each vertex.

1. **Distance from O to A**:
\[ OA = \sqrt{(x + 2)^2 + (y + 3)^2} \]

2. **Distance from O to B**:
\[ OB = \sqrt{(x + 1)^2 + (y - 0)^2} \]

3. **Distance from O to C**:
\[ OC = \sqrt{(x - 7)^2 + (y + 6)^2} \]

### Step 3: Set up the equations based on equal distances
Since OA = OB, we can square both sides to eliminate the square roots:
\[ (x + 2)^2 + (y + 3)^2 = (x + 1)^2 + y^2 \]

Expanding both sides:
\[ (x^2 + 4x + 4 + y^2 + 6y + 9) = (x^2 + 2x + 1 + y^2) \]
Cancelling \(x^2\) and \(y^2\) from both sides:
\[ 4x + 4 + 6y + 9 = 2x + 1 \]
Rearranging gives:
\[ 2x + 6y + 12 = 0 \quad \text{(Equation 1)} \]

Now, we will set up the equation for OB = OC:
\[ (x + 1)^2 + y^2 = (x - 7)^2 + (y + 6)^2 \]

Expanding both sides:
\[ (x^2 + 2x + 1 + y^2) = (x^2 - 14x + 49 + y^2 + 12y + 36) \]
Cancelling \(x^2\) and \(y^2\):
\[ 2x + 1 = -14x + 49 + 12y + 36 \]
Rearranging gives:
\[ 16x - 12y - 84 = 0 \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations
We have two equations:
1. \(2x + 6y + 12 = 0\)
2. \(16x - 12y - 84 = 0\)

We can solve these equations simultaneously. From Equation 1:
\[ 2x + 6y = -12 \implies x + 3y = -6 \quad \text{(Equation 3)} \]

Now, we can express \(x\) in terms of \(y\):
\[ x = -6 - 3y \]

Substituting \(x\) from Equation 3 into Equation 2:
\[ 16(-6 - 3y) - 12y - 84 = 0 \]
\[ -96 - 48y - 12y - 84 = 0 \]
\[ -60y - 180 = 0 \implies y = -3 \]

Now substituting \(y = -3\) back into Equation 3 to find \(x\):
\[ x + 3(-3) = -6 \implies x - 9 = -6 \implies x = 3 \]

Thus, the circumcenter O is at:
\[ O(3, -3) \]

### Step 5: Find the radius of the circumcircle
We can find the radius by calculating the distance from the circumcenter O to any of the triangle's vertices, say A:
\[ OA = \sqrt{(3 - (-2))^2 + (-3 - (-3))^2} = \sqrt{(3 + 2)^2 + (0)^2} = \sqrt{5^2} = 5 \]

### Final Answer
The circumcenter of the triangle is \(O(3, -3)\) and the radius of the circumcircle is \(5\).

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