Show that the four points of intersection of the lines : `(2x-y + 1)` (x-2y+3) = 0`, with the axes lie on a circle and find its centre.

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 show that the four points of intersection of the lines given by the equation \((2x - y + 1)(x - 2y + 3) = 0\) with the axes lie on a circle and find its center.

### Step 1: Find the equations of the lines
The equation \((2x - y + 1)(x - 2y + 3) = 0\) represents two lines:
1. \(2x - y + 1 = 0\)
2. \(x - 2y + 3 = 0\)

### Step 2: Find the intersection points with the axes
**For the first line \(2x - y + 1 = 0\)**:
- **Intersection with the x-axis**: Set \(y = 0\)
\[ 2x + 1 = 0 \implies x = -\frac{1}{2} \]
Point: \((-0.5, 0)\)

- **Intersection with the y-axis**: Set \(x = 0\)
\[ -y + 1 = 0 \implies y = 1 \]
Point: \((0, 1)\)

**For the second line \(x - 2y + 3 = 0\)**:
- **Intersection with the x-axis**: Set \(y = 0\)
\[ x + 3 = 0 \implies x = -3 \]
Point: \((-3, 0)\)

- **Intersection with the y-axis**: Set \(x = 0\)
\[ -2y + 3 = 0 \implies y = \frac{3}{2} \]
Point: \((0, 1.5)\)

### Step 3: List the points of intersection
The four points of intersection with the axes are:
1. \((-0.5, 0)\)
2. \((0, 1)\)
3. \((-3, 0)\)
4. \((0, 1.5)\)

### Step 4: Show that these points lie on a circle
To show that these points lie on a circle, we can use the condition that the general equation of a circle can be expressed in the form:
\[ x^2 + y^2 + Dx + Ey + F = 0 \]
We can also use the condition \(S - \lambda \cdot xy = 0\) where \(S\) is the equation of the circle and \(xy\) represents the axes.

### Step 5: Form the equation
We can express the points in the form of a circle:
1. For the points \((-0.5, 0)\), \((0, 1)\), \((-3, 0)\), and \((0, 1.5)\), we can derive the equation by substituting these points into the general form.

After substituting and simplifying, we arrive at:
\[ 2x^2 + 2y^2 + 7x - 5y + 3 - 5xy = 0 \]

### Step 6: Convert to standard form
Dividing the entire equation by 2 gives:
\[ x^2 + y^2 + \frac{7}{2}x - \frac{5}{2}y + \frac{3}{2} = 0 \]

### Step 7: Identify the center of the circle
Comparing with the standard form of the circle \(x^2 + y^2 + Dx + Ey + F = 0\):
- \(D = \frac{7}{2}\)
- \(E = -\frac{5}{2}\)
- \(F = \frac{3}{2}\)

The center \((h, k)\) of the circle can be found using:
\[ h = -\frac{D}{2} = -\frac{7/2}{2} = -\frac{7}{4} \]
\[ k = -\frac{E}{2} = -\frac{-5/2}{2} = \frac{5}{4} \]

Thus, the center of the circle is:
\[ \left(-\frac{7}{4}, \frac{5}{4}\right) \]

### Final Answer
The four points of intersection lie on a circle, and the center of the circle is \(\left(-\frac{7}{4}, \frac{5}{4}\right)\).