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 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 of the parabola and the circle.

We start with the equations:
1. Parabola: \( x = y^2 \)
2. Circle: \( y^2 + (x - 2)^2 = 8 \)

Substituting \( x = y^2 \) into the circle's equation:
\[ y^2 + (y^2 - 2)^2 = 8 \]
Expanding the equation:
\[ y^2 + (y^4 - 4y^2 + 4) = 8 \]
This simplifies to:
\[ y^4 - 3y^2 - 4 = 0 \]
Let \( z = y^2 \), then we have:
\[ z^2 - 3z - 4 = 0 \]

### Step 2: Solve the quadratic equation.

Using the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ z = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \]
\[ z = \frac{3 \pm \sqrt{9 + 16}}{2} \]
\[ z = \frac{3 \pm 5}{2} \]
Calculating the roots:
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 \( y = \pm 2 \).

### Step 3: Find corresponding x values.

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

### Step 4: Calculate the lengths of the sides of triangle OBC.

1. **Length of BC**:
\[ BC = \sqrt{(4 - 4)^2 + (2 - (-2))^2} = \sqrt{0 + 4^2} = 4 \text{ units} \]

2. **Length of OB**:
\[ OB = \sqrt{(4 - 0)^2 + (2 - 0)^2} = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \text{ units} \]

3. **Length of OC**:
\[ OC = \sqrt{(4 - 0)^2 + (-2 - 0)^2} = \sqrt{4^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \text{ units} \]

### Step 5: Calculate the perimeter of triangle OBC.

The perimeter \( P \) is given by:
\[ P = OB + OC + BC = 2\sqrt{5} + 2\sqrt{5} + 4 = 4 + 4\sqrt{5} \text{ units} \]

Thus, the perimeter of triangle OBC is:
\[ \boxed{4 + 4\sqrt{5}} \text{ units} \]