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.