Show that the points `A(1, 0), B(2, -7), C(8, 1)` and `D(9,-6)` all lie on the same circle. Find the equation of this circle, its centre and radius.

Answer

To show that the points \( A(1, 0) \), \( B(2, -7) \), \( C(8, 1) \), and \( D(9, -6) \) lie on the same circle, we will derive the equations of the circle using the general form and then find the center and radius.

### Step 1: Write the general equation of the circle
The general equation of a circle is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \( (h, k) \) is the center of the circle and \( r \) is the radius.

### Step 2: Substitute each point into the circle equation
We will substitute each point into the equation to form a system of equations.

1. For point \( A(1, 0) \):
\[ (1 - h)^2 + (0 - k)^2 = r^2 \quad \text{(Equation 1)} \]

2. For point \( B(2, -7) \):
\[ (2 - h)^2 + (-7 - k)^2 = r^2 \quad \text{(Equation 2)} \]

3. For point \( C(8, 1) \):
\[ (8 - h)^2 + (1 - k)^2 = r^2 \quad \text{(Equation 3)} \]

4. For point \( D(9, -6) \):
\[ (9 - h)^2 + (-6 - k)^2 = r^2 \quad \text{(Equation 4)} \]

### Step 3: Expand each equation
Now we will expand each equation.

1. **Equation 1**:
\[ (1 - h)^2 + k^2 = r^2 \implies 1 - 2h + h^2 + k^2 = r^2 \]

2. **Equation 2**:
\[ (2 - h)^2 + (-7 - k)^2 = r^2 \implies 4 - 4h + h^2 + 49 + 14k + k^2 = r^2 \]

3. **Equation 3**:
\[ (8 - h)^2 + (1 - k)^2 = r^2 \implies 64 - 16h + h^2 + 1 - 2k + k^2 = r^2 \]

4. **Equation 4**:
\[ (9 - h)^2 + (-6 - k)^2 = r^2 \implies 81 - 18h + h^2 + 36 + 12k + k^2 = r^2 \]

### Step 4: Set the equations equal to each other
Since all equations equal \( r^2 \), we can set them equal to each other to eliminate \( r^2 \).

1. Subtract Equation 1 from Equation 2:
\[ (4 - 4h + 49 + 14k) - (1 - 2h + k^2) = 0 \implies -2h + 14k + 52 = 0 \quad \text{(Equation 5)} \]

2. Subtract Equation 2 from Equation 3:
\[ (64 - 16h + 1 - 2k) - (4 - 4h + 49 + 14k) = 0 \implies -12h - 16k + 12 = 0 \quad \text{(Equation 6)} \]

3. Subtract Equation 3 from Equation 4:
\[ (81 - 18h + 36 + 12k) - (64 - 16h + 1 - 2k) = 0 \implies -2h + 14k + 52 = 0 \quad \text{(Equation 7)} \]

### Step 5: Solve the system of equations
Now we have two equations (Equation 5 and Equation 6) to solve for \( h \) and \( k \).

From Equation 5:
\[ -2h + 14k + 52 = 0 \implies 2h = 14k + 52 \implies h = 7k + 26 \]

Substituting \( h \) into Equation 6:
\[ -12(7k + 26) - 16k + 12 = 0 \]
\[ -84k - 312 - 16k + 12 = 0 \implies -100k - 300 = 0 \implies k = -3 \]

Now substituting \( k = -3 \) back into the equation for \( h \):
\[ h = 7(-3) + 26 = -21 + 26 = 5 \]

### Step 6: Find \( r^2 \)
Now we have \( h = 5 \) and \( k = -3 \). We can substitute these values back into any of the original equations to find \( r^2 \).

Using Equation 1:
\[ (1 - 5)^2 + (0 + 3)^2 = r^2 \implies 16 + 9 = r^2 \implies r^2 = 25 \]

### Step 7: Write the equation of the circle
Thus, the equation of the circle is:
\[ (x - 5)^2 + (y + 3)^2 = 25 \]

### Step 8: Center and radius
The center of the circle is \( (5, -3) \) and the radius is \( r = \sqrt{25} = 5 \).

### Final Result
- **Equation of the circle**: \( (x - 5)^2 + (y + 3)^2 = 25 \)
- **Center**: \( (5, -3) \)
- **Radius**: \( 5 \)