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 can follow these steps:

### Step 1: Write the general equation of the circle
The general equation of a circle can be expressed as:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
where \(g\), \(f\), and \(c\) are constants that we need to determine.

### Step 2: Substitute the first point (2,1)
Substituting the point (2,1) into the equation:
\[ (2)^2 + (1)^2 + 2g(2) + 2f(1) + c = 0 \]
This simplifies to:
\[ 4 + 1 + 4g + 2f + c = 0 \]
Thus, we have:
\[ 4g + 2f + c = -5 \quad \text{(Equation 1)} \]

### Step 3: Substitute the second point (1,2)
Now, substitute the point (1,2):
\[ (1)^2 + (2)^2 + 2g(1) + 2f(2) + c = 0 \]
This simplifies to:
\[ 1 + 4 + 2g + 4f + c = 0 \]
Thus, we have:
\[ 2g + 4f + c = -5 \quad \text{(Equation 2)} \]

### Step 4: Substitute the third point (8,9)
Next, substitute the point (8,9):
\[ (8)^2 + (9)^2 + 2g(8) + 2f(9) + c = 0 \]
This simplifies to:
\[ 64 + 81 + 16g + 18f + c = 0 \]
Thus, we have:
\[ 16g + 18f + c = -145 \quad \text{(Equation 3)} \]

### Step 5: Solve the system of equations
Now we have three equations:
1. \(4g + 2f + c = -5\) (Equation 1)
2. \(2g + 4f + c = -5\) (Equation 2)
3. \(16g + 18f + c = -145\) (Equation 3)

We can eliminate \(c\) by subtracting Equation 1 from Equation 2:
\[ (2g + 4f + c) - (4g + 2f + c) = -5 - (-5) \]
This simplifies to:
\[ -2g + 2f = 0 \quad \Rightarrow \quad f = g \quad \text{(Equation 4)} \]

Now substitute \(f = g\) into Equation 1:
\[ 4g + 2g + c = -5 \quad \Rightarrow \quad 6g + c = -5 \quad \text{(Equation 5)} \]

Now substitute \(f = g\) into Equation 3:
\[ 16g + 18g + c = -145 \quad \Rightarrow \quad 34g + c = -145 \quad \text{(Equation 6)} \]

### Step 6: Solve for \(g\) and \(c\)
Now we can solve Equations 5 and 6:
From Equation 5:
\[ c = -5 - 6g \]
Substituting this into Equation 6:
\[ 34g + (-5 - 6g) = -145 \]
This simplifies to:
\[ 28g - 5 = -145 \quad \Rightarrow \quad 28g = -140 \quad \Rightarrow \quad g = -5 \]

Now substitute \(g = -5\) back into Equation 4:
\[ f = -5 \]

Finally, substitute \(g\) and \(f\) back into Equation 5 to find \(c\):
\[ 6(-5) + c = -5 \quad \Rightarrow \quad -30 + c = -5 \quad \Rightarrow \quad c = 25 \]

### Step 7: Write the final equation of the circle
Now we have:
\[ g = -5, \quad f = -5, \quad c = 25 \]
Substituting these values into the general equation of the circle:
\[ x^2 + y^2 - 10x - 10y + 25 = 0 \]

### Final Answer:
The equation of the circle is:
\[ x^2 + y^2 - 10x - 10y + 25 = 0 \]