The equation of the smallest circle passing from points (1, 1) and (2, 2) and always in the first quadrant is

Options

A. `x^(2)+y^(2)-4x-2y+4=0`
B. `x^(2)+y^(2)+2x+4y+4=0`
C. `x^(2)+y^(2)-3x-3y+4=0`
D. `x^(2)+y^(2)-5x-y+4=0`

Explanation

To find the equation of the smallest circle passing through the points (1, 1) and (2, 2) that is always in the first quadrant, 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.

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

### Step 3: Substitute the second point (2, 2) into the equation
Now, substitute the point (2, 2) into the circle equation:
\[ 2^2 + 2^2 + 2g(2) + 2f(2) + c = 0 \]
This simplifies to:
\[ 8 + 4g + 4f + c = 0 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations
From Equation 1:
\[ c = -2 - 2g - 2f \]
Substituting this expression for \(c\) into Equation 2:
\[ 8 + 4g + 4f - 2 - 2g - 2f = 0 \]
This simplifies to:
\[ 6 + 2g + 2f = 0 \]
Dividing through by 2 gives:
\[ g + f = -3 \quad \text{(Equation 3)} \]

### Step 5: Express \(c\) in terms of \(g\) and \(f\)
Using Equation 3 in the expression for \(c\):
\[ c = -2 - 2g - 2f = -2 - 2(-3) = -2 + 6 = 4 \]

### Step 6: Substitute \(c\) back into Equation 3
From Equation 3:
\[ g + f = -3 \]
We can express \(f\) in terms of \(g\):
\[ f = -3 - g \]

### Step 7: Find the radius of the circle
The radius \(r\) of the circle can be expressed as:
\[ r^2 = g^2 + f^2 - c \]
Substituting \(c = 4\) and \(f = -3 - g\):
\[ r^2 = g^2 + (-3 - g)^2 - 4 \]
Expanding this gives:
\[ r^2 = g^2 + (9 + 6g + g^2) - 4 = 2g^2 + 6g + 5 \]

### Step 8: Minimize \(r^2\)
To minimize \(r^2\), we can complete the square:
\[ r^2 = 2(g^2 + 3g) + 5 = 2\left((g + \frac{3}{2})^2 - \frac{9}{4}\right) + 5 \]
This simplifies to:
\[ r^2 = 2(g + \frac{3}{2})^2 - \frac{9}{2} + 5 = 2(g + \frac{3}{2})^2 + \frac{1}{2} \]
The minimum value occurs when \(g + \frac{3}{2} = 0\) or \(g = -\frac{3}{2}\).

### Step 9: Find \(f\)
Substituting \(g = -\frac{3}{2}\) into \(f = -3 - g\):
\[ f = -3 + \frac{3}{2} = -\frac{3}{2} \]

### Step 10: Write the final equation of the circle
Now substituting \(g\), \(f\), and \(c\) back into the circle equation:
\[ x^2 + y^2 - 3x - 3y + 4 = 0 \]

### Final Answer:
The equation of the smallest circle passing through the points (1, 1) and (2, 2) in the first quadrant is:
\[ x^2 + y^2 - 3x - 3y + 4 = 0 \]