Find the equation to the tangent to the circle `x^(2) + y^(2) = a^(2)` which
is peppendicular to the straight line ` y = mx + c `

Answer

To find the equation of the tangent to the circle \( x^2 + y^2 = a^2 \) that is perpendicular to the line \( y = mx + c \), we can follow these steps:

### Step 1: Identify the Circle and its Properties
The given circle is \( x^2 + y^2 = a^2 \).
- Center: \( (0, 0) \)
- Radius: \( a \)

### Step 2: Slope of the Given Line
The slope of the line \( y = mx + c \) is \( m \).

### Step 3: Find the Slope of the Tangent
Since the tangent is perpendicular to the line, the slope of the tangent line will be the negative reciprocal of \( m \). Thus, the slope \( m_t \) of the tangent is:
\[ m_t = -\frac{1}{m} \]

### Step 4: Equation of the Tangent Line
The general equation of a line with slope \( m_t \) passing through the origin (the center of the circle) can be expressed as:
\[ y = m_t x + k \]
Substituting \( m_t \):
\[ y = -\frac{1}{m} x + k \]

### Step 5: Rearranging the Equation
Rearranging the equation gives:
\[ \frac{1}{m} y + x - k = 0 \]
Multiplying through by \( m \) to eliminate the fraction:
\[ y + mx - mk = 0 \]

### Step 6: Distance from the Center to the Tangent Line
The distance \( d \) from the center \( (0, 0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|C|}{\sqrt{A^2 + B^2}} \]
For our line, \( A = m \), \( B = 1 \), and \( C = -mk \):
\[ d = \frac{|mk|}{\sqrt{m^2 + 1}} \]

### Step 7: Set the Distance Equal to the Radius
Since the tangent touches the circle, this distance must equal the radius \( a \):
\[ \frac{|mk|}{\sqrt{m^2 + 1}} = a \]

### Step 8: Solve for \( k \)
Rearranging gives:
\[ |mk| = a \sqrt{m^2 + 1} \]
Thus, we have:
\[ k = \pm \frac{a \sqrt{m^2 + 1}}{m} \]

### Step 9: Substitute \( k \) Back into the Tangent Equation
Substituting \( k \) back into the tangent line equation gives us two equations:
\[ y + mx - a \sqrt{1 + m^2} = 0 \quad \text{(1)} \]
\[ y + mx + a \sqrt{1 + m^2} = 0 \quad \text{(2)} \]

### Final Answer
The equations of the tangents to the circle \( x^2 + y^2 = a^2 \) that are perpendicular to the line \( y = mx + c \) are:
\[ y + mx - a \sqrt{1 + m^2} = 0 \quad \text{and} \quad y + mx + a \sqrt{1 + m^2} = 0 \]