Find the conditions that the line
(i) y = mx + c may touch the circle `x^(2) +y^(2) = a^(2)` ,
(ii) y = mx + c may touch the circle `x^(2) + y^(2) + 2gx + 2fy + c = 0`.

Answer

To find the conditions under which the line \( y = mx + c \) may touch the given circles, we will analyze each case step by step.

### Part (i): Circle \( x^2 + y^2 = a^2 \)

1. **Identify the Circle's Center and Radius:**
- The equation \( x^2 + y^2 = a^2 \) represents a circle centered at the origin \( (0, 0) \) with a radius \( r = a \).

2. **Find the Perpendicular Distance from the Center to the Line:**
- The line is given by \( y = mx + c \). We can rewrite it in the standard form \( mx - y + c = 0 \).
- The formula for the perpendicular distance \( d \) from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \) is:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
- Here, \( A = m \), \( B = -1 \), \( C = c \), and the point is \( (0, 0) \).
- Substituting these values, we get:
\[ d = \frac{|m(0) - 1(0) + c|}{\sqrt{m^2 + (-1)^2}} = \frac{|c|}{\sqrt{m^2 + 1}} \]

3. **Set the Distance Equal to the Radius:**
- For the line to be tangent to the circle, the distance must equal the radius \( a \):
\[ \frac{|c|}{\sqrt{m^2 + 1}} = a \]

4. **Square Both Sides:**
- Squaring both sides to eliminate the square root gives:
\[ c^2 = a^2(m^2 + 1) \]

5. **Final Condition:**
- Rearranging gives the condition for tangency:
\[ c^2 = a^2 + a^2m^2 \]

### Part (ii): Circle \( x^2 + y^2 + 2gx + 2fy + c = 0 \)

1. **Identify the Circle's Center and Radius:**
- The center of the circle is \( (-g, -f) \) and the radius \( r \) is given by:
\[ r = \sqrt{g^2 + f^2 - c} \]

2. **Find the Perpendicular Distance from the Center to the Line:**
- The line is still \( y = mx + c \), which can be rewritten as \( mx - y + c = 0 \).
- Using the same distance formula, the distance from the center \( (-g, -f) \) to the line is:
\[ d = \frac{|m(-g) - (-f) + c|}{\sqrt{m^2 + 1}} = \frac{|-mg + f + c|}{\sqrt{m^2 + 1}} \]

3. **Set the Distance Equal to the Radius:**
- For the line to be tangent to the circle, we set the distance equal to the radius:
\[ \frac{|-mg + f + c|}{\sqrt{m^2 + 1}} = \sqrt{g^2 + f^2 - c} \]

4. **Square Both Sides:**
- Squaring both sides gives:
\[ (-mg + f + c)^2 = (g^2 + f^2 - c)(m^2 + 1) \]

5. **Final Condition:**
- Expanding both sides will yield the final condition for tangency.