The number of circles that touch all the 3 lines `2x+y=3`, `4x-y=3`, `x+y=2` is

Options

A. 0
B. 2
C. 4
D. infinite

Answer

To solve the problem of finding the number of circles that touch all three lines given by the equations \(2x + y = 3\), \(4x - y = 3\), and \(x + y = 2\), we will follow these steps:

### Step 1: Identify the equations of the lines
We have three lines:
1. \(L_1: 2x + y = 3\)
2. \(L_2: 4x - y = 3\)
3. \(L_3: x + y = 2\)

### Step 2: Find the intersection points of the lines
To find the number of circles that can touch all three lines, we first need to determine the intersection points of these lines.

**Intersection of \(L_1\) and \(L_2\):**
- From \(L_1: y = 3 - 2x\)
- Substitute \(y\) in \(L_2\):
\[ 4x - (3 - 2x) = 3 \]
\[ 4x - 3 + 2x = 3 \]
\[ 6x = 6 \implies x = 1 \]
- Substitute \(x = 1\) back into \(L_1\):
\[ y = 3 - 2(1) = 1 \]
- So, the intersection point of \(L_1\) and \(L_2\) is \((1, 1)\).

**Intersection of \(L_2\) and \(L_3\):**
- From \(L_3: y = 2 - x\)
- Substitute \(y\) in \(L_2\):
\[ 4x - (2 - x) = 3 \]
\[ 4x - 2 + x = 3 \]
\[ 5x = 5 \implies x = 1 \]
- Substitute \(x = 1\) back into \(L_3\):
\[ y = 2 - 1 = 1 \]
- So, the intersection point of \(L_2\) and \(L_3\) is \((1, 1)\).

**Intersection of \(L_1\) and \(L_3\):**
- Substitute \(y\) from \(L_3\) into \(L_1\):
\[ 2x + (2 - x) = 3 \]
\[ 2x + 2 - x = 3 \]
\[ x + 2 = 3 \implies x = 1 \]
- Substitute \(x = 1\) back into \(L_1\):
\[ y = 3 - 2(1) = 1 \]
- So, the intersection point of \(L_1\) and \(L_3\) is \((1, 1)\).

### Step 3: Analyze the intersection points
All three lines intersect at the same point \((1, 1)\). This means that they are concurrent lines.

### Step 4: Determine the number of circles that can touch all three lines
A circle can touch a maximum of two lines at a time if they intersect at a single point. Since all three lines intersect at the same point, it is impossible for a circle to touch all three lines simultaneously.

### Conclusion
Thus, the number of circles that can touch all three lines is:
\[ \text{Number of circles} = 0 \]