Find the equation to the tangent to the circle `x^(2) + y^(2) = a^(2)` which
passes through the point (b,0)

Answer

To find the equation of the tangent to the circle \( x^2 + y^2 = a^2 \) that passes through the point \( (b, 0) \), we can follow these steps:

### Step 1: Write the general equation of the tangent line
The general equation of a line can be expressed in the form:
\[ Ax + By + C = 0 \]
For our tangent line, we will denote it as:
\[ Ax + By + C = 0 \]

### Step 2: Use the point (b, 0) to find C
Since the tangent line passes through the point \( (b, 0) \), we substitute \( x = b \) and \( y = 0 \) into the line equation:
\[ Ab + B(0) + C = 0 \]
This simplifies to:
\[ Ab + C = 0 \implies C = -Ab \]
Thus, the equation of the tangent line can now be written as:
\[ Ax + By - Ab = 0 \]

### Step 3: Calculate the distance from the center of the circle to the line
The center of the circle is at the origin \( (0, 0) \). The distance \( d \) from the center to the line \( Ax + By + C = 0 \) is given by the formula:
\[ d = \frac{|C|}{\sqrt{A^2 + B^2}} \]
Substituting \( C = -Ab \):
\[ d = \frac{|-Ab|}{\sqrt{A^2 + B^2}} = \frac{Ab}{\sqrt{A^2 + B^2}} \]

### Step 4: Set the distance equal to the radius of the circle
Since the line is a tangent to the circle, this distance must equal the radius \( a \) of the circle:
\[ \frac{Ab}{\sqrt{A^2 + B^2}} = a \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \left(\frac{Ab}{\sqrt{A^2 + B^2}}\right)^2 = a^2 \]
This simplifies to:
\[ \frac{A^2b^2}{A^2 + B^2} = a^2 \]

### Step 6: Rearranging the equation
Rearranging the equation gives:
\[ A^2b^2 = a^2(A^2 + B^2) \]
This can be rewritten as:
\[ A^2(b^2 - a^2) = a^2B^2 \]

### Step 7: Solve for B in terms of A
From the equation above, we can express \( B \) in terms of \( A \):
\[ B^2 = \frac{A^2(b^2 - a^2)}{a^2} \]
Taking the square root gives:
\[ B = \pm \frac{A\sqrt{b^2 - a^2}}{a} \]

### Step 8: Substitute B back into the tangent equation
Now we can substitute \( B \) back into the tangent line equation:
\[ Ax + \left(\pm \frac{A\sqrt{b^2 - a^2}}{a}\right)y - Ab = 0 \]
This leads to two possible tangent equations:
\[ Ax + \frac{A\sqrt{b^2 - a^2}}{a}y - Ab = 0 \quad \text{and} \quad Ax - \frac{A\sqrt{b^2 - a^2}}{a}y - Ab = 0 \]

### Final Step: Simplifying the tangent equations
Factoring out \( A \) from both equations gives:
\[ A\left(x + \frac{\sqrt{b^2 - a^2}}{a}y - b\right) = 0 \quad \text{and} \quad A\left(x - \frac{\sqrt{b^2 - a^2}}{a}y - b\right) = 0 \]
Thus, the equations of the tangents are:
\[ x + \frac{\sqrt{b^2 - a^2}}{a}y = b \quad \text{and} \quad x - \frac{\sqrt{b^2 - a^2}}{a}y = b \]