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

\[
(x + 5)^2 + (y + 3)^2 = 80
\]

We are also told that a line is tangent to the circle with a gradient (slope) of 2. We need to find the two possible equations for this line in the form \( y = mx + c \).

### Step 1: Standard form of the equation of a line
The equation of the line is of the form \( y = mx + c \), where \( m = 2 \). Hence, the equation of the line becomes:

\[
y = 2x + c
\]

### Step 2: Equation of the circle
The equation of the circle can be written in the standard form for a circle \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.

From the given equation \( (x + 5)^2 + (y + 3)^2 = 80 \), we can deduce:
- The center of the circle is \( (-5, -3) \).
- The radius \( r = \sqrt{80} = 4\sqrt{5} \).

### Step 3: Condition for tangency
For the line to be tangent to the circle, the perpendicular distance from the center of the circle \((-5, -3)\) to the line \( y = 2x + c \) must equal the radius \( r = 4\sqrt{5} \).

The formula for the perpendicular distance \( d \) from a point \((x_1, y_1)\) to a line \( ax + by + c = 0 \) is:

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

For the line \( y = 2x + c \), we rewrite it as \( 2x - y + c = 0 \). Here, \( A = 2 \), \( B = -1 \), and \( C = c \).

Substitute the center of the circle \( (-5, -3) \) into the formula:

\[
d = \frac{|2(-5) - (-3) + c|}{\sqrt{2^2 + (-1)^2}} = \frac{|-10 + 3 + c|}{\sqrt{5}} = \frac{|c - 7|}{\sqrt{5}}
\]

Since the distance must equal the radius \( r = 4\sqrt{5} \), we set:

\[
\frac{|c - 7|}{\sqrt{5}} = 4\sqrt{5}
\]

### Step 4: Solving for \( c \)
Multiply both sides by \( \sqrt{5} \):

\[
|c - 7| = 20
\]

This gives two possible solutions:

\[
c - 7 = 20 \quad \text{or} \quad c - 7 = -20
\]

Solving these:

1. \( c = 27 \)
2. \( c = -13 \)

### Step 5: Final equations of the tangents
The two possible equations of the tangent lines are:

\[
y = 2x + 27 \quad \text{and} \quad y = 2x - 13
\]

Would you like more details or have any questions on the solution? Here are five related questions to consider:

1. How do you find the equation of a tangent line to a circle geometrically?
2. What would change if the slope of the line were different?
3. How do you derive the formula for the distance from a point to a line?
4. Can there ever be more than two tangents from a point outside the circle to the circle? Why or why not?
5. What if the line was not a tangent but intersected the circle—how would the approach change?

**Tip:** When solving problems involving tangents to circles, always start by checking the distance condition (point to line) to ensure tangency.

The circle C has equation (x + 5)^2 + (y + 3)^2 = 80. The line l is a tangent to the circle and has gradient 2. Find two possible equations for l, giving your answers in the form y = mx + c.

Solve for the two possible equations of the tangent lines to the circle C with center (-5, -3) and radius 4√5. The line has a gradient of 2, and the solution involves using the condition for tangency and the perpendicular distance formula. Suitable for high school students in Grades 10-12.

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