Step 1: Identify the given curves

The two curves are given by the equations:

1. $C_1: y^2 = 4x$ This is the equation of a standard parabola with its vertex at the origin $(0, 0)$ and opening to the right. Its focus is at $(1, 0)$.

2. $C_2: x^2 + y^2 - 6x + 1 = 0$ This is the equation of a circle. To find its center and radius, we rewrite it in the standard form $(x-h)^2 + (y-k)^2 = r^2$. $(x^2 - 6x) + y^2 = -1$ Completing the square for the x-terms: $(x^2 - 6x + 9) + y^2 = -1 + 9$ $(x - 3)^2 + y^2 = 8$ So, $C_2$ is a circle with center $(3, 0)$ and radius $r = \sqrt{8} = 2\sqrt{2}$.

Step 2: Find the points of intersection

To find the points where the curves intersect, we solve their equations simultaneously.

Substitute $y^2 = 4x$ from the equation of $C_1$ into the equation of $C_2$: $x^2 + (4x) - 6x + 1 = 0$ $x^2 - 2x + 1 = 0$ This is a perfect square trinomial: $(x - 1)^2 = 0$ This equation gives a repeated root, $x = 1$.

Now, substitute $x = 1$ back into the equation for the parabola, $y^2 = 4x$, to find the corresponding y-coordinates: $y^2 = 4(1) = 4$ $y = \pm 2$ So, the points of intersection are $(1, 2)$ and $(1, -2)$.

Step 3: Determine the nature of the intersection

The fact that we got a repeated root for $x$ suggests that the curves are tangent at the points of intersection. To verify this, we will find the slopes of the tangents to both curves at these points. If the slopes are equal at a point of intersection, the curves touch each other at that point.

For curve $C_1: y^2 = 4x$ Differentiating with respect to $x$: $2y \frac{dy}{dx} = 4$ $\frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y}$

For curve $C_2: x^2 + y^2 - 6x + 1 = 0$ Differentiating with respect to $x$: $2x + 2y \frac{dy}{dx} - 6 = 0$ $2y \frac{dy}{dx} = 6 - 2x$ $\frac{dy}{dx} = \frac{6 - 2x}{2y} = \frac{3 - x}{y}$

Now, let's check the slopes at the intersection points:

• At the point $(1, 2)$:

  • Slope of tangent to $C_1$: $m_1 = \frac{2}{2} = 1$
  • Slope of tangent to $C_2$: $m_2 = \frac{3 - 1}{2} = \frac{2}{2} = 1$ Since $m_1 = m_2$, the curves have a common tangent at $(1, 2)$ and thus touch each other at this point.

• At the point $(1, -2)$:

  • Slope of tangent to $C_1$: $m_1 = \frac{2}{-2} = -1$
  • Slope of tangent to $C_2$: $m_2 = \frac{3 - 1}{-2} = \frac{2}{-2} = -1$ Since $m_1 = m_2$, the curves also have a common tangent at $(1, -2)$ and touch each other at this point.

Step 4: Conclusion

The curves $C_1$ and $C_2$ touch each other at exactly two distinct points, $(1, 2)$ and $(1, -2)$. Comparing this with the given options:

A: ${C_1}$ and ${C_2}$ touch each other only at one point. (Incorrect) B: ${C_1}$ and ${C_2}$ touch each other exactly at two points. (Correct) C: ${C_1}$ and ${C_2}$ intersect (but do not touch ) at exactly two points. (Incorrect) D: ${C_1}$ and ${C_2}$ neither intersect nor touch each other. (Incorrect)

The correct option is B.