The equation of the smallest circle passing through the point (1,0) and (0,1) is

Options

A. A. ` x^(2) + y^(2) + x + y = 0 `
B. B. ` x^(2) + y^(2) - x - y = 0 `
C. C. ` x^(2) + y^(2) - x - y + 2 = 0 `
D. D. ` x^(2) + y^(2) - x - y - 1 = 0 `

Answer

To find the equation of the smallest circle passing through the points (1,0) and (0,1), we can follow these steps:

### Step 1: Write the general equation of a 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 (1,0) into the equation
Substituting the point (1,0) into the general equation gives:
\[ 1^2 + 0^2 + 2g(1) + 2f(0) + c = 0 \]
This simplifies to:
\[ 1 + 2g + c = 0 \]
From this, we can express \(c\) in terms of \(g\):
\[ c = -1 - 2g \quad \text{(Equation 1)} \]

### Step 3: Substitute the second point (0,1) into the equation
Now, substituting the point (0,1) into the general equation gives:
\[ 0^2 + 1^2 + 2g(0) + 2f(1) + c = 0 \]
This simplifies to:
\[ 1 + 2f + c = 0 \]
From this, we can express \(c\) in terms of \(f\):
\[ c = -1 - 2f \quad \text{(Equation 2)} \]

### Step 4: Set the two expressions for \(c\) equal to each other
From Equations 1 and 2, we have:
\[ -1 - 2g = -1 - 2f \]
This simplifies to:
\[ 2g = 2f \implies g = f \]

### Step 5: Substitute \(g = f\) into one of the equations for \(c\)
Using Equation 1, we substitute \(f\) for \(g\):
\[ c = -1 - 2f \]

### Step 6: Find the radius of the circle
The radius \(r\) of the circle can be expressed as:
\[ r^2 = g^2 + f^2 - c \]
Substituting \(c\) gives:
\[ r^2 = g^2 + g^2 - (-1 - 2g) = 2g^2 + 1 + 2g \]

### Step 7: Differentiate \(r^2\) with respect to \(g\)
To find the minimum radius, we differentiate:
\[ \frac{dr^2}{dg} = 4g + 2 \]
Setting this equal to zero for minimization:
\[ 4g + 2 = 0 \implies g = -\frac{1}{2} \]

### Step 8: Substitute \(g\) back to find \(f\) and \(c\)
Since \(g = f\):
\[ f = -\frac{1}{2} \]
Now substituting \(g\) into the equation for \(c\):
\[ c = -1 - 2(-\frac{1}{2}) = -1 + 1 = 0 \]

### Step 9: Write the final equation of the circle
Now substituting \(g\), \(f\), and \(c\) back into the general equation:
\[ x^2 + y^2 + 2(-\frac{1}{2})x + 2(-\frac{1}{2})y + 0 = 0 \]
This simplifies to:
\[ x^2 + y^2 - x - y = 0 \]

### Final Answer
The equation of the smallest circle passing through the points (1,0) and (0,1) is:
\[ x^2 + y^2 - x - y = 0 \]