The circle `x^2+y^2=5` meets the parabola `y^2=4x` at `P` and `Q` . Then the length `P Q` is equal to (a)2 (b) `2sqrt(2)` (c) 4 (d) none of these

Options

A. 2
B. `2sqrt(2)`
C. 4
D. none of these

Explanation

To find the length \( PQ \) where the circle \( x^2 + y^2 = 5 \) meets the parabola \( y^2 = 4x \), we will follow these steps:

### Step 1: Substitute the equation of the parabola into the circle's equation.
The equation of the parabola is given by:
\[ y^2 = 4x \]
Substituting \( y^2 \) from the parabola into the circle's equation \( x^2 + y^2 = 5 \):
\[ x^2 + 4x = 5 \]

### Step 2: Rearrange the equation.
Rearranging the equation gives:
\[ x^2 + 4x - 5 = 0 \]

### Step 3: Solve the quadratic equation.
Now, we will solve the quadratic equation \( x^2 + 4x - 5 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 4, c = -5 \):
\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-5)}}{2 \cdot 1} \]
\[ x = \frac{-4 \pm \sqrt{16 + 20}}{2} \]
\[ x = \frac{-4 \pm \sqrt{36}}{2} \]
\[ x = \frac{-4 \pm 6}{2} \]
Calculating the two possible values for \( x \):
1. \( x = \frac{2}{2} = 1 \)
2. \( x = \frac{-10}{2} = -5 \)

### Step 4: Determine the corresponding \( y \) values.
Now, we will find the corresponding \( y \) values using \( y^2 = 4x \):
1. For \( x = 1 \):
\[ y^2 = 4 \cdot 1 = 4 \implies y = \pm 2 \]
So, the points are \( P(1, 2) \) and \( Q(1, -2) \).

2. For \( x = -5 \):
\[ y^2 = 4 \cdot (-5) = -20 \quad \text{(not possible)} \]
Thus, we discard \( x = -5 \).

### Step 5: Calculate the length \( PQ \).
The length \( PQ \) is given by the distance between the points \( P(1, 2) \) and \( Q(1, -2) \):
\[ PQ = |y_2 - y_1| = |(-2) - (2)| = |-4| = 4 \]

### Conclusion:
Thus, the length \( PQ \) is equal to \( 4 \).

### Final Answer:
(c) 4

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