If `P(1,2sqrt(2)),R(9,0), S(-1,0)`, then radius of the circumcircle of `DeltaPRS,` is

Options

A. 5
B. `3sqrt3`
C. `3sqrt2`
D. `2sqrt3`

Explanation

To find the radius of the circumcircle of triangle \( \Delta PRS \) with points \( P(1, 2\sqrt{2}) \), \( R(9, 0) \), and \( S(-1, 0) \), we will follow these steps:

### Step 1: Set up the equation of the circumcircle
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 we need to determine.

### Step 2: Substitute point \( P(1, 2\sqrt{2}) \)
Substituting the coordinates of point \( P(1, 2\sqrt{2}) \) into the circle equation:
\[ 1^2 + (2\sqrt{2})^2 + 2g(1) + 2f(2\sqrt{2}) + c = 0 \]
This simplifies to:
\[ 1 + 8 + 2g + 4\sqrt{2}f + c = 0 \]
Thus, we have:
\[ 2g + 4\sqrt{2}f + c + 9 = 0 \quad \text{(Equation 1)} \]

### Step 3: Substitute point \( R(9, 0) \)
Now substituting the coordinates of point \( R(9, 0) \):
\[ 9^2 + 0^2 + 2g(9) + 2f(0) + c = 0 \]
This simplifies to:
\[ 81 + 18g + c = 0 \quad \text{(Equation 2)} \]

### Step 4: Substitute point \( S(-1, 0) \)
Next, substituting the coordinates of point \( S(-1, 0) \):
\[ (-1)^2 + 0^2 + 2g(-1) + 2f(0) + c = 0 \]
This simplifies to:
\[ 1 - 2g + c = 0 \quad \text{(Equation 3)} \]

### Step 5: Solve the system of equations
We now have three equations:
1. \( 2g + 4\sqrt{2}f + c + 9 = 0 \)
2. \( 81 + 18g + c = 0 \)
3. \( 1 - 2g + c = 0 \)

From Equation 3, we can express \( c \):
\[ c = 2g - 1 \quad \text{(Substituting into Equations 1 and 2)} \]

Substituting \( c \) into Equation 2:
\[ 81 + 18g + (2g - 1) = 0 \implies 20g + 80 = 0 \implies g = -4 \]

Now substituting \( g = -4 \) into Equation 3:
\[ c = 2(-4) - 1 = -8 - 1 = -9 \]

Substituting \( g = -4 \) into Equation 1 to find \( f \):
\[ 2(-4) + 4\sqrt{2}f - 9 + 9 = 0 \implies -8 + 4\sqrt{2}f = 0 \implies 4\sqrt{2}f = 8 \implies f = \sqrt{2} \]

### Step 6: Calculate the radius of the circumcircle
The radius \( R \) of the circumcircle is given by:
\[ R = \sqrt{g^2 + f^2 - c} \]
Substituting the values of \( g \), \( f \), and \( c \):
\[ R = \sqrt{(-4)^2 + (\sqrt{2})^2 - (-9)} = \sqrt{16 + 2 + 9} = \sqrt{27} = 3\sqrt{3} \]

### Final Answer
The radius of the circumcircle of triangle \( \Delta PRS \) is:
\[ \boxed{3\sqrt{3}} \]