The radius of the circle
` sqrt(1 + a^(2)) (x^(2) + y^(2)) - 2bx - 2aby = 0 ` is

Options

A. b
B. a pair of lines
C. ab
D. `sqrt(1 + a^(2))`

Answer

To find the radius of the circle given by the equation

\[ \sqrt{1 + a^{2}} (x^{2} + y^{2}) - 2bx - 2aby = 0, \]

we can follow these steps:

### Step 1: Rearranging the equation
First, we can rearrange the equation to isolate the terms involving \(x\) and \(y\):

\[ \sqrt{1 + a^{2}} (x^{2} + y^{2}) = 2bx + 2aby. \]

### Step 2: Dividing by \(\sqrt{1 + a^{2}}\)
Next, we divide both sides by \(\sqrt{1 + a^{2}}\) to simplify the equation:

\[ x^{2} + y^{2} = \frac{2bx}{\sqrt{1 + a^{2}}} + \frac{2aby}{\sqrt{1 + a^{2}}}. \]

### Step 3: Rearranging into standard form
Now, we can rearrange this into the standard form of a circle equation:

\[ x^{2} + y^{2} - \frac{2b}{\sqrt{1 + a^{2}}}x - \frac{2ab}{\sqrt{1 + a^{2}}}y = 0. \]

### Step 4: Completing the square
To find the center and radius, we complete the square for both \(x\) and \(y\):

1. For \(x\):
\[ x^{2} - \frac{2b}{\sqrt{1 + a^{2}}}x = \left(x - \frac{b}{\sqrt{1 + a^{2}}}\right)^{2} - \left(\frac{b}{\sqrt{1 + a^{2}}}\right)^{2}. \]

2. For \(y\):
\[ y^{2} - \frac{2ab}{\sqrt{1 + a^{2}}}y = \left(y - \frac{ab}{\sqrt{1 + a^{2}}}\right)^{2} - \left(\frac{ab}{\sqrt{1 + a^{2}}}\right)^{2}. \]

### Step 5: Putting it all together
Substituting these back into the equation gives us:

\[ \left(x - \frac{b}{\sqrt{1 + a^{2}}}\right)^{2} + \left(y - \frac{ab}{\sqrt{1 + a^{2}}}\right)^{2} = \left(\frac{b^{2}}{1 + a^{2}} + \frac{a^{2}b^{2}}{1 + a^{2}}\right). \]

### Step 6: Finding the radius
The right side simplifies to:

\[ \frac{b^{2}(1 + a^{2})}{1 + a^{2}} = \frac{b^{2}}{1 + a^{2}}. \]

Thus, the radius \(r\) of the circle is given by:

\[ r = \sqrt{\frac{b^{2}}{1 + a^{2}}} = \frac{b}{\sqrt{1 + a^{2}}}. \]

### Final Answer
The radius of the circle is

\[ \frac{b}{\sqrt{1 + a^{2}}}. \]

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