Intersection and Tangency of Parabola and Circle

To determine the relationship between the two curves, we analyze their points of intersection and their tangents at those points.

The given curves are:
$C_1 : y^2 = 4x$
$C_2 : x^2 + y^2 - 6x + 1 = 0$

Step 1: Find the Points of Intersection

Substitute $y^2 = 4x$ from equation $C_1$ into the equation of $C_2$:
$x^2 + 4x - 6x + 1 = 0$
$x^2 - 2x + 1 = 0$
$(x - 1)^2 = 0 \implies x = 1$

Since $x = 1$, we find the corresponding values of $y$ using $C_1$:
$y^2 = 4(1) = 4 \implies y = \pm 2$

Thus, the two curves intersect at two points:
$P(1, 2) \quad \text{and} \quad Q(1, -2)$

Step 2: Check the Slopes of the Tangents at the Points of Intersection

For the parabola $C_1$:
$y^2 = 4x$
Differentiating both sides with respect to $x$:
$2y \frac{dy}{dx} = 4 \implies \left(\frac{dy}{dx}\right)_{C_1} = \frac{2}{y}$

For the circle $C_2$:
$x^2 + y^2 - 6x + 1 = 0$
Differentiating both sides with respect to $x$:
$2x + 2y \frac{dy}{dx} - 6 = 0 \implies \left(\frac{dy}{dx}\right)_{C_2} = \frac{3 - x}{y}$

At the point $P(1, 2)$:

• 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 share a common tangent at $P(1, 2)$, meaning they touch each other at this point.

At the point $Q(1, -2)$:

• 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 share a common tangent at $Q(1, -2)$, meaning they touch each other at this point as well.

Conclusion:

$C_1$ and $C_2$ touch each other exactly at two points.

Therefore, the correct option is B.