Find the polar equation of a circle, the initial line being a tangent. What does it become if the origin be on the circumference?

Answer

To find the polar equation of a circle with the initial line being a tangent, we can follow these steps:

### Step 1: Understand the Circle's Position
The circle is tangent to the initial line (the polar axis, which is the x-axis in Cartesian coordinates) at a point. Let's denote the radius of the circle as \( a \) and the angle of the tangent line from the origin as \( \alpha \).

### Step 2: Determine the Center of the Circle
The center of the circle will be at a distance \( a \cot \alpha \) from the origin along the line making an angle \( \alpha \) with the polar axis. The coordinates of the center in Cartesian form will be:
- \( x = a \cot \alpha \)
- \( y = a \)

### Step 3: Write the Cartesian Equation of the Circle
The standard form of the equation of a circle in Cartesian coordinates is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \( (h, k) \) is the center and \( r \) is the radius. Substituting the values we have:
\[ (x - a \cot \alpha)^2 + (y - a)^2 = a^2 \]

### Step 4: Convert to Polar Coordinates
To convert this equation to polar coordinates, we use the substitutions:
- \( x = r \cos \theta \)
- \( y = r \sin \theta \)

Substituting these into the circle's equation gives:
\[ (r \cos \theta - a \cot \alpha)^2 + (r \sin \theta - a)^2 = a^2 \]

### Step 5: Expand and Simplify
Expanding both sides:
1. For \( (r \cos \theta - a \cot \alpha)^2 \):
\[ r^2 \cos^2 \theta - 2a \cot \alpha r \cos \theta + a^2 \cot^2 \alpha \]
2. For \( (r \sin \theta - a)^2 \):
\[ r^2 \sin^2 \theta - 2ar \sin \theta + a^2 \]

Combining these:
\[ r^2 (\cos^2 \theta + \sin^2 \theta) - 2ar (\sin \theta + \cot \alpha \cos \theta) + (a^2 \cot^2 \alpha + a^2) = a^2 \]

### Step 6: Use the Identity \( \cos^2 \theta + \sin^2 \theta = 1 \)
This simplifies to:
\[ r^2 - 2ar (\sin \theta + \cot \alpha \cos \theta) + a^2 \cot^2 \alpha = 0 \]

### Step 7: Rearranging the Equation
Rearranging gives:
\[ r^2 - 2ar \left(\sin \theta + \frac{\cos \theta}{\sin \alpha}\right) + a^2 = 0 \]

### Step 8: Identify the Polar Equation
This is the general form of the polar equation of a circle with a tangent at the initial line.

### Step 9: Case When the Origin is on the Circumference
If the origin is on the circumference of the circle, then the distance from the center to the origin is equal to the radius \( a \). In this case, \( a \cot \alpha = 0 \) implies that \( \alpha = 90^\circ \) (the tangent line is vertical).

Substituting \( \alpha = 90^\circ \) into the polar equation simplifies it to:
\[ r = 2a \sin \theta \]

### Final Answers
1. The polar equation of the circle when the initial line is a tangent is:
\[ r^2 - 2ar \left(\sin \theta + \cot \alpha \cos \theta\right) + a^2 = 0 \]
2. If the origin is on the circumference, the equation becomes:
\[ r = 2a \sin \theta \]