If the 4 points made by intersection of lines `2x-y+1=0, x-2y+3=0` with the coordinate axes are concylic then centre of circle is

Options

A. (7/4,5/4)
B. (7/4,-5/4)
C. (-7/4,5/4)
D. (-7/4,-5/4)

Answer

To solve the problem, we need to find the center of the circle that passes through the four points formed by the intersection of the lines with the coordinate axes. Let's break down the solution step by step.

### Step 1: Find the intersection points of the lines with the coordinate axes.

**For the first line:** \(2x - y + 1 = 0\)

1. **Find the x-intercept (set \(y = 0\)):**
\[ 2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2} \]
So, the point is \(A\left(-\frac{1}{2}, 0\right)\).

2. **Find the y-intercept (set \(x = 0\)):**
\[ -y + 1 = 0 \implies y = 1 \]
So, the point is \(B(0, 1)\).

**For the second line:** \(x - 2y + 3 = 0\)

1. **Find the x-intercept (set \(y = 0\)):**
\[ x + 3 = 0 \implies x = -3 \]
So, the point is \(C(-3, 0)\).

2. **Find the y-intercept (set \(x = 0\)):**
\[ -2y + 3 = 0 \implies 2y = 3 \implies y = \frac{3}{2} \]
So, the point is \(D\left(0, \frac{3}{2}\right)\).

### Summary of Points
- \(A\left(-\frac{1}{2}, 0\right)\)
- \(B(0, 1)\)
- \(C(-3, 0)\)
- \(D\left(0, \frac{3}{2}\right)\)

### Step 2: Check if the points are concyclic.
The points are concyclic if the distances from the center of the circle to each of the points are equal. Let's denote the center of the circle as \(C(h, k)\).

### Step 3: Set up the equations for distances.
1. Distance from \(C\) to \(A\):
\[ CA^2 = (h + \frac{1}{2})^2 + (k - 0)^2 \]

2. Distance from \(C\) to \(B\):
\[ CB^2 = (h - 0)^2 + (k - 1)^2 \]

3. Distance from \(C\) to \(C\):
\[ CC^2 = (h + 3)^2 + (k - 0)^2 \]

4. Distance from \(C\) to \(D\):
\[ CD^2 = (h - 0)^2 + (k - \frac{3}{2})^2 \]

### Step 4: Set the distances equal.
We can set \(CA^2 = CB^2\):
\[ (h + \frac{1}{2})^2 + k^2 = h^2 + (k - 1)^2 \]

Expanding both sides:
\[ h^2 + h + \frac{1}{4} + k^2 = h^2 + k^2 - 2k + 1 \]
Cancelling \(h^2\) and \(k^2\):
\[ h + \frac{1}{4} = -2k + 1 \]
Rearranging gives:
\[ h + 2k = \frac{3}{4} \quad \text{(Equation 1)} \]

Now set \(CA^2 = CC^2\):
\[ (h + \frac{1}{2})^2 + k^2 = (h + 3)^2 + k^2 \]
Cancelling \(k^2\):
\[ (h + \frac{1}{2})^2 = (h + 3)^2 \]
Expanding gives:
\[ h^2 + h + \frac{1}{4} = h^2 + 6h + 9 \]
Cancelling \(h^2\):
\[ h + \frac{1}{4} = 6h + 9 \]
Rearranging gives:
\[ -5h = \frac{35}{4} \implies h = -\frac{7}{4} \quad \text{(Equation 2)} \]

### Step 5: Substitute \(h\) into Equation 1 to find \(k\).
Substituting \(h = -\frac{7}{4}\) into Equation 1:
\[ -\frac{7}{4} + 2k = \frac{3}{4} \]
Solving for \(k\):
\[ 2k = \frac{3}{4} + \frac{7}{4} = \frac{10}{4} \implies k = \frac{5}{4} \]

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