The two circles `x^(2)+y^(2)-cx=0 and x^(2)+y^(2)=4` touch each other if:

Options

A. `|c|=4`
B. `|c|=2`
C. `|c|=8`
D. None of these

Explanation

To determine the condition under which the two circles \( x^2 + y^2 - cx = 0 \) and \( x^2 + y^2 = 4 \) touch each other, we will follow these steps:

### Step 1: Rewrite the equations of the circles
The first circle can be rewritten as:
\[ x^2 + y^2 = cx \]
This can be rearranged to:
\[ x^2 + y^2 = \frac{c^2}{4} \]
To see this, we can complete the square:
\[ x^2 - cx + y^2 = 0 \implies (x - \frac{c}{2})^2 + y^2 = \frac{c^2}{4} \]
Thus, the center of the first circle \( C_1 \) is \( \left( \frac{c}{2}, 0 \right) \) and its radius \( r_1 \) is \( \frac{c}{2} \).

The second circle is already in standard form:
\[ x^2 + y^2 = 4 \]
Thus, the center of the second circle \( C_2 \) is \( (0, 0) \) and its radius \( r_2 \) is \( 2 \).

### Step 2: Calculate the distance between the centers
The distance \( d \) between the centers \( C_1 \) and \( C_2 \) is given by:
\[ d = \sqrt{\left( \frac{c}{2} - 0 \right)^2 + (0 - 0)^2} = \frac{|c|}{2} \]

### Step 3: Set up the conditions for tangency
The circles touch each other if either of the following conditions holds:
1. They touch externally:
\[ d = r_1 + r_2 \]
This gives us:
\[ \frac{|c|}{2} = \frac{c}{2} + 2 \]

2. They touch internally:
\[ d = |r_1 - r_2| \]
This gives us:
\[ \frac{|c|}{2} = \left| \frac{c}{2} - 2 \right| \]

### Step 4: Solve the equations

**For external tangency:**
\[ \frac{|c|}{2} = \frac{c}{2} + 2 \]
This implies:
\[ |c| = c + 4 \]
- If \( c \geq 0 \):
\[ c = c + 4 \implies 0 = 4 \quad \text{(no solution)} \]
- If \( c < 0 \):
\[ -c = c + 4 \implies -2c = 4 \implies c = -2 \]

**For internal tangency:**
\[ \frac{|c|}{2} = \left| \frac{c}{2} - 2 \right| \]
- If \( c \geq 0 \):
\[ \frac{c}{2} = \frac{c}{2} - 2 \implies 0 = -2 \quad \text{(no solution)} \]
- If \( c < 0 \):
\[ -\frac{c}{2} = \frac{c}{2} - 2 \implies -c = c - 4 \implies -2c = -4 \implies c = 2 \quad \text{(not valid as \( c < 0 \))} \]

### Conclusion
The only valid solution for \( c \) is:
\[ c = -2 \]
Thus, the two circles touch each other if \( |c| = 2 \).