The four point of intersection of the lines `( 2x -y +1) ( x- 2y +3) = 0` with the axes lie on a circle whose center centre is at the point `:`

Options

A. `( -7//4, 5//4)`
B. `( 3//4,5//4)`
C. ` ( 9//4,5//4)`
D. `(0,5//4)`

Answer

To solve the problem, we need to find the center of the circle that passes through the four points of intersection of the lines given by the equation \((2x - y + 1)(x - 2y + 3) = 0\) with the axes.

### Step-by-Step Solution:

1. **Identify the Lines**:
The equation \((2x - y + 1)(x - 2y + 3) = 0\) represents two lines:
- Line 1: \(2x - y + 1 = 0\)
- Line 2: \(x - 2y + 3 = 0\)

2. **Find Intercepts of Line 1**:
To find the intercepts of Line 1, we can rewrite it in slope-intercept form:
\[ y = 2x + 1 \]
- **X-Intercept**: Set \(y = 0\):
\[ 0 = 2x + 1 \implies x = -\frac{1}{2} \quad \text{(Point A: } (-\frac{1}{2}, 0)\text{)} \]
- **Y-Intercept**: Set \(x = 0\):
\[ y = 1 \quad \text{(Point B: } (0, 1)\text{)} \]

3. **Find Intercepts of Line 2**:
Rewrite Line 2 in slope-intercept form:
\[ 2y = x + 3 \implies y = \frac{1}{2}x + \frac{3}{2} \]
- **X-Intercept**: Set \(y = 0\):
\[ 0 = \frac{1}{2}x + \frac{3}{2} \implies x = -3 \quad \text{(Point C: } (-3, 0)\text{)} \]
- **Y-Intercept**: Set \(x = 0\):
\[ y = \frac{3}{2} \quad \text{(Point D: } (0, \frac{3}{2})\text{)} \]

4. **List the Points of Intersection**:
The four points of intersection with the axes are:
- A: \((- \frac{1}{2}, 0)\)
- B: \((0, 1)\)
- C: \((-3, 0)\)
- D: \((0, \frac{3}{2})\)

5. **Equation of the Circle**:
The circle passing through these points can be determined using the general form of a circle:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
We can find the coefficients by substituting the points into this equation.

6. **Substituting Points**:
Substitute each point into the circle equation:
- For Point A \((- \frac{1}{2}, 0)\):
\[ \left(-\frac{1}{2}\right)^2 + 0^2 + 2g\left(-\frac{1}{2}\right) + 2f(0) + c = 0 \implies \frac{1}{4} - g + c = 0 \]
- For Point B \((0, 1)\):
\[ 0^2 + 1^2 + 2g(0) + 2f(1) + c = 0 \implies 1 + 2f + c = 0 \]
- For Point C \((-3, 0)\):
\[ (-3)^2 + 0^2 + 2g(-3) + 2f(0) + c = 0 \implies 9 - 6g + c = 0 \]
- For Point D \((0, \frac{3}{2})\):
\[ 0^2 + \left(\frac{3}{2}\right)^2 + 2g(0) + 2f\left(\frac{3}{2}\right) + c = 0 \implies \frac{9}{4} + 3f + c = 0 \]

7. **Solving the System of Equations**:
We now have a system of equations to solve for \(g\), \(f\), and \(c\):
1. \(\frac{1}{4} - g + c = 0\)
2. \(1 + 2f + c = 0\)
3. \(9 - 6g + c = 0\)
4. \(\frac{9}{4} + 3f + c = 0\)

By solving these equations, we can find the values of \(g\) and \(f\).

8. **Finding the Center**:
The center of the circle \((h, k)\) is given by:
\[ h = -g, \quad k = -f \]

### Final Answer:
After solving the equations, we find that the center of the circle is at the point:
\[ \left(-\frac{7}{4}, \frac{5}{4}\right) \]