This problem is an excellent test of your understanding of the properties of parabolas and circles, and how to determine conditions for tangency between curves. We will systematically break down each part of the problem.

1. Analyze the First Parabola and Determine its Key Features

• Key Concept: The standard form of a parabola opening to the right with its vertex at the origin is $y^2 = 4ax$.

  • Its vertex is at $(0,0)$.
  • Its focus is at $(a,0)$.

• Step-by-Step Working:

  1. The given parabola is $y^2 = 2x$.
  2. We compare this to the standard form $y^2 = 4ax$.
  3. By comparing coefficients, we have $4a = 2$.
  4. Solving for $a$, we get $a = \frac{2}{4} = \frac{1}{2}$.

• Explanation: We identify the parameter 'a' because it directly gives us the coordinates of the focus and helps in determining the vertex.

  • The vertex of the parabola $y^2 = 2x$ is $V = (0,0)$.
  • The focus of the parabola $y^2 = 2x$ is $F = (a,0) = \left(\frac{1}{2}, 0\right)$.

2. Determine the Equation of the Circle

• Key Concept: The equation of a circle with center $(h,k)$ and radius $R$ is $(x-h)^2 + (y-k)^2 = R^2$.

  • The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.

• Step-by-Step Working:

  1. The circle has a radius $R = 2$. Therefore, $R^2 = 4$.
  2. The circle passes through the vertex $V=(0,0)$ and the focus $F=\left(\frac{1}{2}, 0\right)$ of the first parabola.
  3. Let the center of the circle be $(h,k)$.
  4. Since the circle passes through $V(0,0)$, the distance from $(h,k)$ to $(0,0)$ must be equal to the radius $R$. $(h-0)^2 + (k-0)^2 = R^2$ $h^2 + k^2 = 2^2 = 4 \quad \text{(Equation 1)}$
  5. Since the circle passes through $F\left(\frac{1}{2}, 0\right)$, the distance from $(h,k)$ to $\left(\frac{1}{2}, 0\right)$ must also be equal to the radius $R$. $\left(h-\frac{1}{2}\right)^2 + (k-0)^2 = R^2$ $\left(h-\frac{1}{2}\right)^2 + k^2 = 4 \quad \text{(Equation 2)}$
  6. Now, we solve Equations 1 and 2 for $h$ and $k$. Substitute $k^2 = 4 - h^2$ from Equation 1 into Equation 2: $\left(h-\frac{1}{2}\right)^2 + (4 - h^2) = 4$ $$$$