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Let's solve this step by step.

### Problem:
The circle \( C \) has the equation:
\[
x^2 + y^2 - 4x - 6 = 0
\]
The line \( l \) has the equation:
\[
y = 3x - 6
\]

We'll address each sub-question.

---

#### (a) **Find the center of the circle**:
1. Start with the given equation of the circle:
   \[
   x^2 + y^2 - 4x - 6 = 0
   \]
2. Rewrite it in the standard form by completing the square:
   - Group \( x \)-terms: \( x^2 - 4x \).
   - Complete the square: \( x^2 - 4x \) becomes \( (x - 2)^2 - 4 \).
   - Substitute this back:
     \[
     (x - 2)^2 + y^2 - 4 - 6 = 0
     \]
   - Simplify:
     \[
     (x - 2)^2 + y^2 = 10
     \]
3. This is the standard form of a circle:
   \[
   (x - h)^2 + (y - k)^2 = r^2
   \]
   Here, \( (h, k) = (2, 0) \) is the center, and \( r = \sqrt{10} \) is the radius.

**Answer:** The center of the circle is \( (2, 0) \).

---

#### (b) **Show that the line \( l \) passes through the center**:
The center of the circle is \( (2, 0) \), and the equation of the line is \( y = 3x - 6 \). To check if the line passes through the center, substitute \( x = 2 \) into the equation of the line:
\[
y = 3(2) - 6 = 6 - 6 = 0
\]
Since \( (2, 0) \) satisfies the line's equation, the line passes through the center of the circle.

**Answer:** The line \( l \) passes through the center \( (2, 0) \).

---

#### (c) **Sketch a diagram to show both \( C \) and \( l \)**:
I can provide a textual description for the sketch or generate it if you'd like! Here's the structure of the diagram:
1. The circle is centered at \( (2, 0) \) with radius \( \sqrt{10} \).
2. The line \( l \) passes through \( (2, 0) \) with slope 3, so it slants upward at a steep angle.

Would you like me to generate this diagram for you?

---

#### (d) **Find equations for the tangents to \( C \) that are parallel to \( l \)**:
The slope of line \( l \) is 3. For tangents parallel to \( l \), their slope must also be 3.

1. Start with the general equation of a line with slope 3:
   \[
   y = 3x + c
   \]
   Here \( c \) is the y-intercept.

2. For a line to be tangent to the circle, the perpendicular distance from the center \( (2, 0) \) to the line must equal the radius \( \sqrt{10} \). The perpendicular distance from \( (h, k) \) to a line \( Ax + By + C = 0 \) is:
   \[
   d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}
   \]
   Substituting \( A = -3 \), \( B = 1 \), \( C = -c \), and the center \( (2, 0) \):
   \[
   d = \frac{|(-3)(2) + (0)(1) - c|}{\sqrt{(-3)^2 + 1^2}} = \sqrt{10}
   \]
   Simplify:
   \[
   \frac{| -6 - c |}{\sqrt{10}} = \sqrt{10}
   \]
3. Multiply through by \( \sqrt{10} \):
   \[
   | -6 - c | = 10
   \]
   Solve for \( c \):
   \[
   -6 - c = \pm 10
   \]
   Case 1: \( -6 - c = 10 \) gives \( c = -16 \).  
   Case 2: \( -6 - c = -10 \) gives \( c = 4 \).

Thus, the equations of the tangents are:
\[
y = 3x - 16 \quad \text{and} \quad y = 3x + 4
\]

**Answer:** The tangents parallel to \( l \) are:
\[
y = 3x - 16 \quad \text{and} \quad y = 3x + 4.
\]

---

Would you like me to:
1. Generate the sketch for part (c)?
2. Explain any step further?  
3. Clarify concepts like completing the square or tangent distances?  

Here are five related questions to explore:
1. How do you derive the standard form of a circle from its general equation?  
2. Why is the perpendicular distance used for finding tangent lines?  
3. How would the solution change if the circle's radius were smaller or larger?  
4. Can two tangents to a circle ever intersect?  
5. How do slopes of parallel lines affect their equations?

**Tip:** For equations of tangents, always link their slope to properties of the curve or circle!

The circle C has equation x^2 + y^2 - 4x - 6 = 0 and the line l has equation y = 3x - 6. Find the center of the circle, verify if the line passes through it, sketch the circle and line, and find equations for tangents to the circle parallel to the line.

Solve a detailed geometry problem involving the equation of a circle, the equation of a line, and tangents parallel to the line. Concepts include completing the square, verifying line equations, and using the perpendicular distance formula. Suitable for high school students.

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