Let B and C are the points of intersection of the parabola `x=y^(2)` and the circle `y^(2)+(x-2)^(2)=8`. The perimeter (in units) of the triangle OBC, where O is the origin, is

Options

A. 8
B. `4sqrt5`
C. `4sqrt5+2`
D. `4(sqrt5+1)`

Answer

To find the perimeter of triangle OBC, where O is the origin (0,0), and B and C are the points of intersection of the parabola \(x = y^2\) and the circle \(y^2 + (x - 2)^2 = 8\), we will follow these steps:

### Step 1: Find the points of intersection
We start with the two equations:
1. \(x = y^2\) (Equation of the parabola)
2. \(y^2 + (x - 2)^2 = 8\) (Equation of the circle)

Substituting \(x = y^2\) into the second equation:

\[ y^2 + (y^2 - 2)^2 = 8 \]

### Step 2: Simplify the equation
Expanding the equation:

\[ y^2 + (y^4 - 4y^2 + 4) = 8 \]

Combining like terms:

\[ y^4 - 3y^2 - 4 = 0 \]

### Step 3: Let \(z = y^2\)
Let \(z = y^2\). Then the equation becomes:

\[ z^2 - 3z - 4 = 0 \]

### Step 4: Solve the quadratic equation
Using the quadratic formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

Here, \(a = 1\), \(b = -3\), and \(c = -4\):

\[ z = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \]
\[ z = \frac{3 \pm \sqrt{9 + 16}}{2} \]
\[ z = \frac{3 \pm 5}{2} \]

Calculating the two possible values for \(z\):

1. \(z = \frac{8}{2} = 4\)
2. \(z = \frac{-2}{2} = -1\) (not valid since \(z = y^2\) cannot be negative)

Thus, \(y^2 = 4\) implies \(y = 2\) or \(y = -2\).

### Step 5: Find corresponding \(x\) values
Using \(y^2 = 4\):

- For \(y = 2\), \(x = y^2 = 4\) → Point C(4, 2)
- For \(y = -2\), \(x = y^2 = 4\) → Point B(4, -2)

### Step 6: Calculate the lengths of the sides of triangle OBC
Now we have the points:
- O(0, 0)
- B(4, -2)
- C(4, 2)

#### Length OB:
\[ OB = \sqrt{(4 - 0)^2 + (-2 - 0)^2} = \sqrt{4^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \]

#### Length OC:
\[ OC = \sqrt{(4 - 0)^2 + (2 - 0)^2} = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \]

#### Length BC:
\[ BC = \sqrt{(4 - 4)^2 + (2 - (-2))^2} = \sqrt{0 + (2 + 2)^2} = \sqrt{16} = 4 \]

### Step 7: Calculate the perimeter of triangle OBC
The perimeter \(P\) is given by:

\[ P = OB + OC + BC = 2\sqrt{5} + 2\sqrt{5} + 4 = 4\sqrt{5} + 4 \]

### Final Answer:
Thus, the perimeter of triangle OBC is:

\[ \boxed{4(1 + \sqrt{5})} \]