Let `P(alpha,beta)` be a point in the first quadrant. Circles are drawn through P touching the coordinate axes.
Radius of one of the circles is

Options

A. `(sqrt(a)-sqrt(beta))^(2)`
B. `(sqrt(alpha)+sqrt(beta))^(2)`
C. `alpha +beta -sqrt(alpha beta)`
D. `alpha +beta -sqrt(2alpha beta)`

Explanation

To find the radius of the circle that passes through the point \( P(\alpha, \beta) \) and touches the coordinate axes, we can follow these steps:

### Step 1: Understand the Circle's Properties
Since the circle touches the coordinate axes, the center of the circle must be at the point \( (r, r) \), where \( r \) is the radius of the circle. This is because the distance from the center to both axes must equal the radius.

### Step 2: Write the Equation of the Circle
The general form of the equation of a circle with center \( (h, k) \) and radius \( r \) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
For our case, substituting \( h = r \) and \( k = r \):
\[ (x - r)^2 + (y - r)^2 = r^2 \]

### Step 3: Expand the Equation
Expanding the equation gives:
\[ (x^2 - 2rx + r^2) + (y^2 - 2ry + r^2) = r^2 \]
This simplifies to:
\[ x^2 + y^2 - 2rx - 2ry + 2r^2 = r^2 \]
Rearranging gives:
\[ x^2 + y^2 - 2rx - 2ry + r^2 = 0 \]

### Step 4: Substitute the Point \( P(\alpha, \beta) \)
Since the circle passes through the point \( P(\alpha, \beta) \), we substitute \( x = \alpha \) and \( y = \beta \):
\[ \alpha^2 + \beta^2 - 2r\alpha - 2r\beta + r^2 = 0 \]

### Step 5: Rearranging the Equation
Rearranging the equation gives:
\[ \alpha^2 + \beta^2 + r^2 = 2r(\alpha + \beta) \]

### Step 6: Isolate \( r \)
To isolate \( r \), we can rearrange it as follows:
\[ r^2 - 2r(\alpha + \beta) + (\alpha^2 + \beta^2) = 0 \]

### Step 7: Use the Quadratic Formula
This is a quadratic equation in \( r \). We can use the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where:
- \( a = 1 \)
- \( b = -2(\alpha + \beta) \)
- \( c = \alpha^2 + \beta^2 \)

Substituting these values into the quadratic formula:
\[ r = \frac{2(\alpha + \beta) \pm \sqrt{(-2(\alpha + \beta))^2 - 4(\alpha^2 + \beta^2)}}{2} \]

### Step 8: Simplify the Expression
Calculating the discriminant:
\[ (-2(\alpha + \beta))^2 = 4(\alpha + \beta)^2 \]
And:
\[ -4(\alpha^2 + \beta^2) \]
Thus, the discriminant becomes:
\[ 4(\alpha + \beta)^2 - 4(\alpha^2 + \beta^2) = 4((\alpha + \beta)^2 - (\alpha^2 + \beta^2)) \]
This simplifies to:
\[ 4(\alpha^2 + 2\alpha\beta + \beta^2 - \alpha^2 - \beta^2) = 8\alpha\beta \]
So, we have:
\[ r = \frac{2(\alpha + \beta) \pm \sqrt{8\alpha\beta}}{2} \]
This simplifies to:
\[ r = (\alpha + \beta) \pm 2\sqrt{2\alpha\beta} \]

### Step 9: Choose the Correct Radius
Since we are looking for a positive radius, we take:
\[ r = \alpha + \beta - 2\sqrt{2\alpha\beta} \]

### Final Answer
Thus, the radius of the circle is:
\[ \boxed{\alpha + \beta - 2\sqrt{2\alpha\beta}} \]