Length of the common chord of the parabola `y^(2)=8x` and the circle `x^(2)+y^(2)-2x-4y=0` is :

Options

A. `(1)/(2)sqrt5`
B. `sqrt5`
C. `2sqrt5`
D. `3sqrt5`

Answer

To find the length of the common chord of the parabola \( y^2 = 8x \) and the circle \( x^2 + y^2 - 2x - 4y = 0 \), we will follow these steps:

### Step 1: Rewrite the Circle Equation
First, we will rewrite the equation of the circle in standard form. The given equation is:

\[ x^2 + y^2 - 2x - 4y = 0 \]

We can rearrange it as follows:

\[ x^2 - 2x + y^2 - 4y = 0 \]

Now, we complete the square for both \(x\) and \(y\):

- For \(x^2 - 2x\), we add and subtract \(1\) (which is \((\frac{-2}{2})^2\)):
\[ x^2 - 2x + 1 = (x - 1)^2 \]

- For \(y^2 - 4y\), we add and subtract \(4\) (which is \((\frac{-4}{2})^2\)):
\[ y^2 - 4y + 4 = (y - 2)^2 \]

Thus, we can rewrite the circle equation as:

\[ (x - 1)^2 + (y - 2)^2 = 5 \]

This shows that the circle has a center at \((1, 2)\) and a radius of \(\sqrt{5}\).

### Step 2: Parametrize the Parabola
The parabola \(y^2 = 8x\) can be parametrized using the parameter \(t\):

\[ x = 2t^2, \quad y = 4t \]

### Step 3: Substitute into the Circle Equation
Next, we substitute the parametric equations of the parabola into the equation of the circle:

\[ (2t^2 - 1)^2 + (4t - 2)^2 = 5 \]

Expanding this:

1. For \((2t^2 - 1)^2\):
\[ (2t^2 - 1)^2 = 4t^4 - 4t^2 + 1 \]

2. For \((4t - 2)^2\):
\[ (4t - 2)^2 = 16t^2 - 16t + 4 \]

Combining these, we have:

\[ 4t^4 - 4t^2 + 1 + 16t^2 - 16t + 4 = 5 \]

This simplifies to:

\[ 4t^4 + 12t^2 - 16t + 5 - 5 = 0 \]

Thus, we have:

\[ 4t^4 + 12t^2 - 16t = 0 \]

### Step 4: Factor the Equation
We can factor out \(t\):

\[ t(4t^3 + 12t - 16) = 0 \]

This gives us one solution \(t = 0\). Now we need to solve the cubic equation:

\[ 4t^3 + 12t - 16 = 0 \]

### Step 5: Find Roots of the Cubic Equation
Using the Rational Root Theorem or synthetic division, we can find that \(t = 1\) is a root. Dividing \(4t^3 + 12t - 16\) by \(t - 1\) gives us:

\[ 4t^2 + 4t + 16 = 0 \]

The discriminant of this quadratic is:

\[ D = 4^2 - 4 \cdot 4 \cdot 4 = 16 - 64 = -48 < 0 \]

Thus, the only real solutions are \(t = 0\) and \(t = 1\).

### Step 6: Find Points of Intersection
Now we find the points corresponding to \(t = 0\) and \(t = 1\):

- For \(t = 0\):
\[ (x, y) = (2(0)^2, 4(0)) = (0, 0) \]

- For \(t = 1\):
\[ (x, y) = (2(1)^2, 4(1)) = (2, 4) \]

### Step 7: Calculate Length of the Common Chord
The length of the common chord is the distance between the points \((0, 0)\) and \((2, 4)\):

Using the distance formula:

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

### Final Answer
The length of the common chord is \(2\sqrt{5}\).

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