Consider two curves
`C_1:y^2 = 4x, C_z : x^2 + y^2 – 6x+1=0` then

Options

A. `C_1 and C_2` touch each other only at one point
B. `C_1 and C_2` touch each other exactly at two points
C. `C_1 and C_2` intersect (but do not touch) at exactly two points.
D. `C_1 and C_2` neither intersect nor touch each other

Answer

To solve the problem, we need to analyze the two given curves:

1. **Curve C1**: \( y^2 = 4x \) (This is a parabola that opens to the right.)
2. **Curve C2**: \( x^2 + y^2 - 6x + 1 = 0 \) (This is a circle.)

### Step 1: Rewrite Curve C2
First, we rewrite the equation of the circle in standard form.

Starting with:
\[ x^2 + y^2 - 6x + 1 = 0 \]

We can complete the square for the \( x \) terms:
\[ x^2 - 6x + y^2 + 1 = 0 \]
\[ (x^2 - 6x + 9) + y^2 - 9 + 1 = 0 \]
\[ (x - 3)^2 + y^2 = 8 \]

This represents a circle centered at \( (3, 0) \) with a radius of \( \sqrt{8} \).

### Step 2: Substitute \( y^2 \) from C1 into C2
Now, we substitute \( y^2 \) from Curve C1 into the equation of Curve C2.

From Curve C1:
\[ y^2 = 4x \]

Substituting this into Curve C2:
\[ (x - 3)^2 + 4x = 8 \]

### Step 3: Simplify the equation
Now, we simplify the equation:
\[ (x - 3)^2 + 4x - 8 = 0 \]
Expanding \( (x - 3)^2 \):
\[ x^2 - 6x + 9 + 4x - 8 = 0 \]
Combining like terms:
\[ x^2 - 2x + 1 = 0 \]

### Step 4: Factor the quadratic equation
The equation can be factored as:
\[ (x - 1)^2 = 0 \]

### Step 5: Find the roots
Setting the factored equation to zero gives:
\[ x - 1 = 0 \implies x = 1 \]

### Step 6: Find corresponding \( y \) values
Now, we substitute \( x = 1 \) back into the equation of Curve C1 to find the corresponding \( y \) values:
\[ y^2 = 4(1) = 4 \implies y = \pm 2 \]

### Conclusion
The two curves \( C_1 \) and \( C_2 \) touch each other at the points \( (1, 2) \) and \( (1, -2) \).

### Final Answer
The curves \( C_1 \) and \( C_2 \) touch each other at exactly two points: \( (1, 2) \) and \( (1, -2) \).

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