The equation of a circle passing through the vertex and the extremites of the latus rectum of the parabola `y ^(2)=8x` is

Options

A. `x^(2) + y^(2) + 10x =0`
B. `x ^(2) + y^(2) + 10 y =0`
C. `x^(2) + y ^(2) -10 x=0`
D. `x ^(2) + y ^(2) -5x =0`

Answer

To find the equation of the circle passing through the vertex and the extremities of the latus rectum of the parabola \( y^2 = 8x \), we will follow these steps:

### Step 1: Identify the vertex of the parabola
The given parabola \( y^2 = 8x \) is in the standard form \( y^2 = 4px \), where \( p \) is the distance from the vertex to the focus. Here, \( 4p = 8 \), so \( p = 2 \).

The vertex of the parabola is at the origin:
\[ V(0, 0) \]

### Step 2: Find the focus of the parabola
The focus of the parabola is located at \( (p, 0) \). Since \( p = 2 \), the focus is at:
\[ F(2, 0) \]

### Step 3: Determine the extremities of the latus rectum
The latus rectum of the parabola is a line segment perpendicular to the axis of symmetry of the parabola that passes through the focus. The length of the latus rectum is given by \( 4p \).

For our parabola, the length of the latus rectum is:
\[ 4p = 4 \times 2 = 8 \]

The extremities of the latus rectum can be found by moving \( 4 \) units up and down from the focus. Therefore, the coordinates of the extremities are:
\[ L_1(2, 4) \quad \text{and} \quad L_2(2, -4) \]

### Step 4: Use the points to find the equation of the circle
The circle must pass through the three points: the vertex \( V(0, 0) \), and the extremities of the latus rectum \( L_1(2, 4) \) and \( L_2(2, -4) \).

The general equation of a circle is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \( (h, k) \) is the center of the circle and \( r \) is its radius.

### Step 5: Set up equations using the points
1. For the vertex \( (0, 0) \):
\[ (0 - h)^2 + (0 - k)^2 = r^2 \quad \Rightarrow \quad h^2 + k^2 = r^2 \quad \text{(1)} \]

2. For the point \( (2, 4) \):
\[ (2 - h)^2 + (4 - k)^2 = r^2 \quad \Rightarrow \quad (2 - h)^2 + (4 - k)^2 = r^2 \quad \text{(2)} \]

3. For the point \( (2, -4) \):
\[ (2 - h)^2 + (-4 - k)^2 = r^2 \quad \Rightarrow \quad (2 - h)^2 + (-4 - k)^2 = r^2 \quad \text{(3)} \]

### Step 6: Solve the equations
From equations (2) and (3), we can set them equal to each other since they both equal \( r^2 \):
\[ (2 - h)^2 + (4 - k)^2 = (2 - h)^2 + (-4 - k)^2 \]

Expanding both sides:
\[ (4 - k)^2 = (-4 - k)^2 \]

This simplifies to:
\[ 16 - 8k + k^2 = 16 + 8k + k^2 \]

Cancelling \( k^2 \) and \( 16 \) from both sides gives:
\[ -8k = 8k \quad \Rightarrow \quad 16k = 0 \quad \Rightarrow \quad k = 0 \]

Substituting \( k = 0 \) back into equation (1):
\[ h^2 + 0^2 = r^2 \quad \Rightarrow \quad h^2 = r^2 \quad \Rightarrow \quad r = |h| \]

Substituting \( k = 0 \) into equation (2):
\[ (2 - h)^2 + 4^2 = r^2 \quad \Rightarrow \quad (2 - h)^2 + 16 = h^2 \]

Expanding and simplifying:
\[ 4 - 4h + h^2 + 16 = h^2 \quad \Rightarrow \quad 20 - 4h = 0 \quad \Rightarrow \quad h = 5 \]

Thus, the center of the circle is \( (5, 0) \) and the radius is \( r = 5 \).

### Step 7: Write the final equation of the circle
The equation of the circle is:
\[ (x - 5)^2 + y^2 = 25 \]

### Final Answer
The equation of the circle is:
\[ \boxed{(x - 5)^2 + y^2 = 25} \]