The radius of the circle `r^(2)-2sqrt(2r) (cos theta + sin theta)-5=0`, is

Options

A. 9
B. 5
C. 3
D. 2

Answer

To find the radius of the circle given by the equation \( r^2 - 2\sqrt{2}r(\cos \theta + \sin \theta) - 5 = 0 \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ r^2 - 2\sqrt{2}r(\cos \theta + \sin \theta) - 5 = 0 \]
This equation can be rearranged to express it in a more useful form.

### Step 2: Substitute for \(\cos \theta\) and \(\sin \theta\)
Using the relationships \( \cos \theta = \frac{x}{r} \) and \( \sin \theta = \frac{y}{r} \), we substitute these into the equation:
\[ r^2 - 2\sqrt{2}r\left(\frac{x}{r} + \frac{y}{r}\right) - 5 = 0 \]
This simplifies to:
\[ r^2 - 2\sqrt{2}(x + y) - 5 = 0 \]

### Step 3: Isolate \(r^2\)
Now we can isolate \(r^2\):
\[ r^2 = 2\sqrt{2}(x + y) + 5 \]

### Step 4: Use the relationship between \(x\), \(y\), and \(r\)
From the Pythagorean theorem, we know:
\[ x^2 + y^2 = r^2 \]
Substituting \(r^2\) from the previous step:
\[ x^2 + y^2 = 2\sqrt{2}(x + y) + 5 \]

### Step 5: Rearranging to standard form
Rearranging gives us:
\[ x^2 + y^2 - 2\sqrt{2}(x + y) - 5 = 0 \]

### Step 6: Identify the center and radius
This equation can be compared to the standard form of a circle:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
To convert it into this form, we complete the square for both \(x\) and \(y\).

1. For \(x\):
\[ x^2 - 2\sqrt{2}x \rightarrow (x - \sqrt{2})^2 - 2 \]
2. For \(y\):
\[ y^2 - 2\sqrt{2}y \rightarrow (y - \sqrt{2})^2 - 2 \]

Putting these together:
\[ (x - \sqrt{2})^2 + (y - \sqrt{2})^2 - 2 - 2 - 5 = 0 \]
This simplifies to:
\[ (x - \sqrt{2})^2 + (y - \sqrt{2})^2 = 9 \]

### Step 7: Find the radius
From the equation \((x - \sqrt{2})^2 + (y - \sqrt{2})^2 = 9\), we see that the radius \(r\) is:
\[ r = \sqrt{9} = 3 \]

### Final Answer
Thus, the radius of the circle is \(3\).

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