Find the equation of circle passing through the point `(2,1), (1,2)` and` (8,9)`.

Answer

To find the equation of the circle passing through the points (2, 1), (1, 2), and (8, 9), we will use the general equation of a circle, which is given by:

\[ x^2 + y^2 + 2Gx + 2Fy + C = 0 \]

where \( G \), \( F \), and \( C \) are constants that we need to determine.

### Step 1: Substitute the first point (2, 1)

Substituting \( x = 2 \) and \( y = 1 \) into the equation:

\[ 2^2 + 1^2 + 2G(2) + 2F(1) + C = 0 \]

Calculating this gives:

\[ 4 + 1 + 4G + 2F + C = 0 \]

This simplifies to:

\[ 5 + 4G + 2F + C = 0 \quad \text{(Equation 1)} \]

### Step 2: Substitute the second point (1, 2)

Now, substituting \( x = 1 \) and \( y = 2 \):

\[ 1^2 + 2^2 + 2G(1) + 2F(2) + C = 0 \]

Calculating this gives:

\[ 1 + 4 + 2G + 4F + C = 0 \]

This simplifies to:

\[ 5 + 2G + 4F + C = 0 \quad \text{(Equation 2)} \]

### Step 3: Substitute the third point (8, 9)

Next, substituting \( x = 8 \) and \( y = 9 \):

\[ 8^2 + 9^2 + 2G(8) + 2F(9) + C = 0 \]

Calculating this gives:

\[ 64 + 81 + 16G + 18F + C = 0 \]

This simplifies to:

\[ 145 + 16G + 18F + C = 0 \quad \text{(Equation 3)} \]

### Step 4: Solve the system of equations

Now we have three equations:

1. \( 5 + 4G + 2F + C = 0 \) (Equation 1)
2. \( 5 + 2G + 4F + C = 0 \) (Equation 2)
3. \( 145 + 16G + 18F + C = 0 \) (Equation 3)

We can eliminate \( C \) by subtracting Equation 1 from Equation 2:

\[ (5 + 2G + 4F + C) - (5 + 4G + 2F + C) = 0 \]

This simplifies to:

\[ -2G + 2F = 0 \implies G = F \quad \text{(Equation 4)} \]

### Step 5: Substitute \( F \) in terms of \( G \)

Now substitute \( F = G \) into Equation 1:

\[ 5 + 4G + 2G + C = 0 \implies 5 + 6G + C = 0 \implies C = -5 - 6G \quad \text{(Equation 5)} \]

### Step 6: Substitute \( G \) and \( C \) into Equation 3

Now substitute \( F = G \) and \( C = -5 - 6G \) into Equation 3:

\[ 145 + 16G + 18G - 5 - 6G = 0 \]

This simplifies to:

\[ 140 + 28G = 0 \implies G = -5 \]

### Step 7: Find \( F \) and \( C \)

Since \( F = G \):

\[ F = -5 \]

Now substitute \( G \) into Equation 5 to find \( C \):

\[ C = -5 - 6(-5) = -5 + 30 = 25 \]

### Step 8: Write the final equation of the circle

Now substituting \( G \), \( F \), and \( C \) back into the general equation of the circle:

\[ x^2 + y^2 + 2(-5)x + 2(-5)y + 25 = 0 \]

This simplifies to:

\[ x^2 + y^2 - 10x - 10y + 25 = 0 \]

Thus, the equation of the circle is:

\[ \boxed{x^2 + y^2 - 10x - 10y + 25 = 0} \]