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The given problem describes a circle \( C \) with equation:

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

This is the standard form of a circle equation, where the center of the circle is at \( (-5, -3) \), and the radius \( r \) is:

\[
r = \sqrt{80} = 4\sqrt{5}
\]

The line \( l \) is a tangent to the circle, with the gradient \( m = 2 \). We are asked to find two possible equations for \( l \) in the form \( y = mx + c \).

### Step-by-Step Solution:

1. **Equation of the Tangent Line**: 
   The general equation of a line is \( y = mx + c \), where \( m \) is the gradient (given as 2), and \( c \) is the y-intercept. Therefore, we know the general form for our tangent lines will be:

   \[
   y = 2x + c
   \]

2. **Condition for Tangency**:
   A line is tangent to a circle if the perpendicular distance from the center of the circle to the line is equal to the radius of the circle.

   The formula for the perpendicular distance \( d \) from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \) is:

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

   In this case, the line is \( 2x - y + c = 0 \) (rearranged form), and the center of the circle is \( (-5, -3) \). Substituting these into the distance formula:

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

3. **Equating Distance to the Radius**:
   Since the line is a tangent, the distance \( d \) must be equal to the radius of the circle, which is \( 4\sqrt{5} \):

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

   Multiply both sides by \( \sqrt{5} \):

   \[
   | -7 + c | = 20
   \]

4. **Solve for \( c \)**:
   The absolute value equation gives us two cases:

   - Case 1: \( -7 + c = 20 \), which gives \( c = 27 \)
   - Case 2: \( -7 + c = -20 \), which gives \( c = -13 \)

5. **Equations of the Tangent Lines**:
   Therefore, the two possible equations for the tangent lines are:

   \[
   y = 2x + 27
   \]
   \[
   y = 2x - 13
   \]

These are the two required equations for the tangent lines to the circle.

---

Would you like more details or have any questions?

### Follow-up Questions:
1. How do you find the center and radius of a circle from its equation?
2. What is the condition for a line to be tangent to a circle?
3. How do you use the distance formula for a point and a line?
4. What are the steps to solve an absolute value equation?
5. How can you verify that a line is tangent to a circle?

### Tip:
For a quick check if a line is tangent to a circle, compute the discriminant of the quadratic equation formed by substituting the line into the circle equation. If it is zero, the line is tangent!

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.

Learn how to find the equations of two possible tangent lines to the circle (x + 5)^2 + (y + 3)^2 = 80 with gradient 2. This problem involves using the equation of a circle and applying the distance formula for tangency conditions. Suitable for students in Grades 11-12.

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