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To solve this problem, we need to find the two possible values of \( m \) for which the line \( y = mx - 2 \) is tangent to the circle given by the equation:

\[
x^2 + 6x + y^2 - 8y + 5 = 0.
\]

### Step 1: Write the equation of the circle in standard form
We can complete the square for both the \(x\)-terms and the \(y\)-terms.

#### For \(x^2 + 6x\):
Complete the square:
\[
x^2 + 6x = (x + 3)^2 - 9.
\]

#### For \(y^2 - 8y\):
Complete the square:
\[
y^2 - 8y = (y - 4)^2 - 16.
\]

Now substitute these into the original equation:
\[
(x + 3)^2 - 9 + (y - 4)^2 - 16 + 5 = 0.
\]
Simplifying:
\[
(x + 3)^2 + (y - 4)^2 - 20 = 0.
\]
\[
(x + 3)^2 + (y - 4)^2 = 20.
\]
This is the equation of a circle with center \( (-3, 4) \) and radius \( \sqrt{20} = 2\sqrt{5} \).

### Step 2: Use the condition for tangency
The line \( y = mx - 2 \) can be rewritten as:
\[
mx - y = -2.
\]
The general formula for the distance from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \) is:
\[
d = \frac{|Ax_1 + By_1

The line y = mx - 2 is a tangent to the circle x^2 + 6x + y^2 - 8y + 5 = 0. Find the two possible values of m, giving your answers in exact form.

Solve for the slope 'm' of the tangent line y = mx - 2 that touches the circle x^2 + 6x + y^2 - 8y + 5 = 0. This problem uses the distance formula and the tangency condition in coordinate geometry. Suitable for Grades 10-12.

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