From the focus of the parabola `y^2=2px` as centre, a circle is drawn so that a common chord of the curve is equidistant from the vertex and the focus. The radius of the circle is

Options

A. `(p)/(4)`
B. `(3p)/(4)`
C. `(3p)/(8)`
D. `(sqrt(3)p)/(8)`

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Identify the focus of the parabola
The equation of the parabola is given as \( y^2 = 2px \). We can compare this with the standard form \( y^2 = 4ax \) to find the value of \( a \).

- From \( 2p = 4a \), we have:
\[ a = \frac{p}{2} \]

- The focus of the parabola is located at \( (a, 0) = \left(\frac{p}{2}, 0\right) \).

### Step 2: Define the center of the circle
The center of the circle is at the focus of the parabola, which we found to be \( \left(\frac{p}{2}, 0\right) \).

### Step 3: Understand the condition for the common chord
We need to find a common chord of the parabola and the circle such that it is equidistant from the vertex and the focus. The vertex of the parabola is at \( (0, 0) \), and the focus is at \( \left(\frac{p}{2}, 0\right) \).

Let’s denote the point where the chord intersects the x-axis as \( P \) with coordinates \( (x, y) \).

### Step 4: Set up the distance condition
According to the problem, the distance from the vertex \( V(0, 0) \) to the chord must equal the distance from the focus \( F\left(\frac{p}{2}, 0\right) \) to the chord.

Let \( P \) be the point on the chord where it intersects the x-axis. The distances can be expressed as:
- Distance from \( V \) to \( P \): \( VP = x \)
- Distance from \( F \) to \( P \): \( PF = \left| x - \frac{p}{2} \right| \)

Setting these distances equal gives:
\[ x = \left| x - \frac{p}{2} \right| \]

### Step 5: Solve the distance equation
This equation can be solved in two cases:

**Case 1:** \( x - \frac{p}{2} \geq 0 \) (i.e., \( x \geq \frac{p}{2} \))
\[ x = x - \frac{p}{2} \implies \frac{p}{2} = 0 \quad \text{(not possible)} \]

**Case 2:** \( x - \frac{p}{2} < 0 \) (i.e., \( x < \frac{p}{2} \))
\[ x = -\left(x - \frac{p}{2}\right) \implies x = -x + \frac{p}{2} \implies 2x = \frac{p}{2} \implies x = \frac{p}{4} \]

### Step 6: Find the y-coordinate of the intersection point
Now, substitute \( x = \frac{p}{4} \) into the parabola's equation to find the corresponding \( y \)-coordinates:
\[ y^2 = 2p\left(\frac{p}{4}\right) = \frac{p^2}{2} \implies y = \pm \frac{p}{\sqrt{2}} \]

### Step 7: Equation of the circle
The equation of the circle centered at \( \left(\frac{p}{2}, 0\right) \) with radius \( r \) is:
\[ \left(x - \frac{p}{2}\right)^2 + y^2 = r^2 \]

### Step 8: Substitute the point into the circle's equation
Substituting \( \left(\frac{p}{4}, \frac{p}{\sqrt{2}}\right) \) into the circle's equation gives:
\[ \left(\frac{p}{4} - \frac{p}{2}\right)^2 + \left(\frac{p}{\sqrt{2}}\right)^2 = r^2 \]

Calculating the left side:
\[ \left(-\frac{p}{4}\right)^2 + \frac{p^2}{2} = \frac{p^2}{16} + \frac{p^2}{2} = \frac{p^2}{16} + \frac{8p^2}{16} = \frac{9p^2}{16} \]

Thus, we have:
\[ r^2 = \frac{9p^2}{16} \implies r = \frac{3p}{4} \]

### Conclusion
The radius of the circle is:
\[ \boxed{\frac{3p}{4}} \]