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We are given the equation of the circle \(C\) as:

\[
(x - 6)^2 + (y - 5)^2 = 17
\]

This represents a circle with a center at \((6, 5)\) and a radius \( \sqrt{17} \).

The lines \(l_1\) and \(l_2\) are tangents to this circle, and both intersect at the point \((0, 12)\), which means they pass through this point.

### Step 1: General Equation of a Tangent Line
The equation of a line passing through a point \((x_1, y_1)\) and having a slope \(m\) is:

\[
y - y_1 = m(x - x_1)
\]

Here, the point is \((0, 12)\), so the equation for any line passing through this point becomes:

\[
y - 12 = m(x - 0) \implies y = mx + 12
\]

So, both \(l_1\) and \(l_2\) can be written in the form \(y = mx + 12\), but we need to determine the slopes \(m_1\) and \(m_2\) for each of these lines.

### Step 2: Condition for Tangency
For a line to be tangent to a circle, the perpendicular distance from the center of the circle to the line must equal the radius of the circle. The distance \(d\) from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\) is given by:

\[
d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}
\]

Here, the equation of the line \(y = mx + 12\) can be rewritten as:

\[
mx - y + 12 = 0
\]

In this form, \(A = m\), \(B = -1\), and \(C = 12\). The center of the circle is \((6, 5)\), and the radius is \(\sqrt{17}\). Using the formula for the distance from a point to a line:

\[
\frac{|m(6) - (5) + 12|}{\sqrt{m^2 + 1}} = \sqrt{17}
\]

Simplifying the numerator:

\[
|6m + 7| = \sqrt{17} \cdot \sqrt{m^2 + 1}
\]

Square both sides to remove the square roots:

\[
(6m + 7)^2 = 17(m^2 + 1)
\]

Expand both sides:

\[
36m^2 + 84m + 49 = 17m^2 + 17
\]

Simplify:

\[
19m^2 + 84m + 32 = 0
\]

### Step 3: Solve the Quadratic Equation
Solve the quadratic equation using the quadratic formula:

\[
m = \frac{-84 \pm \sqrt{84^2 - 4(19)(32)}}{2(19)}
\]

First, calculate the discriminant:

\[
84^2 = 7056, \quad 4 \times 19 \times 32 = 2432
\]

\[
\sqrt{7056 - 2432} = \sqrt{4624} = 68
\]

Now, solve for \(m\):

\[
m = \frac{-84 \pm 68}{38}
\]

This gives two solutions:

\[
m_1 = \frac{-84 + 68}{38} = \frac{-16}{38} = -\frac{8}{19}, \quad m_2 = \frac{-84 - 68}{38} = \frac{-152}{38} = -4
\]

### Step 4: Equations of the Tangent Lines
Now that we have the slopes, the equations of the tangent lines \(l_1\) and \(l_2\) are:

For \(l_1\):

\[
y = -\frac{8}{19}x + 12
\]

For \(l_2\):

\[
y = -4x + 12
\]

### Final Answer:
The equations of the tangent lines are:

- \(l_1: y = -\frac{8}{19}x + 12\)
- \(l_2: y = -4x + 12\)

Would you like more details or have any questions?

Here are some related questions to expand on this problem:
1. How would you find the equation of a tangent to a circle at a specific point on the circle?
2. How does the discriminant of a quadratic equation help in understanding the nature of solutions?
3. How would the solution change if the circle had a different center?
4. Can a line intersect a circle at more than two points? Why or why not?
5. How does the slope of the tangent relate to the radius at the point of tangency?

**Tip:** Always make sure to check if the solution makes sense by verifying key conditions, such as distances or specific points being on the lines.

The circle C has equation (x - 6)^2 + (y - 5)^2 = 17. The lines l1 and l2 are each a tangent to the circle and intersect at the point (0, 12). Find the equations of l1 and l2, giving your answers in the form y = mx + c.

This problem involves finding the equations of two tangent lines to a circle that intersect at the point (0, 12). Key concepts include the equation of tangents, solving quadratic equations, and applying the condition for tangency. Suitable for students in Grades 10-12.

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