If `2x+3y-6=0` and `9x+6y-18=0` cuts the axes in concyclic points, then the centre of the circle, is

Options

A. (2, 3)
B. (3, 2)
C. (5, 5)
D. (5/2, 5/2)

Answer

To solve the problem, we need to find the center of the circle that passes through the points where the lines \(2x + 3y - 6 = 0\) and \(9x + 6y - 18 = 0\) intersect the axes.

### Step 1: Find the points where the first line intersects the axes.

The first line is given by the equation:
\[ 2x + 3y - 6 = 0 \]

**Finding the y-intercept (set \(x = 0\)):**
\[ 2(0) + 3y - 6 = 0 \implies 3y = 6 \implies y = 2 \]
So, the y-intercept is at the point \((0, 2)\).

**Finding the x-intercept (set \(y = 0\)):**
\[ 2x + 3(0) - 6 = 0 \implies 2x = 6 \implies x = 3 \]
So, the x-intercept is at the point \((3, 0)\).

### Step 2: Find the points where the second line intersects the axes.

The second line is given by the equation:
\[ 9x + 6y - 18 = 0 \]

**Finding the y-intercept (set \(x = 0\)):**
\[ 9(0) + 6y - 18 = 0 \implies 6y = 18 \implies y = 3 \]
So, the y-intercept is at the point \((0, 3)\).

**Finding the x-intercept (set \(y = 0\)):**
\[ 9x + 6(0) - 18 = 0 \implies 9x = 18 \implies x = 2 \]
So, the x-intercept is at the point \((2, 0)\).

### Step 3: List the points of intersection.

From the calculations, we have the following points:
- From the first line: \((0, 2)\) and \((3, 0)\)
- From the second line: \((0, 3)\) and \((2, 0)\)

### Step 4: Check if these points are concyclic.

The points are \((0, 2)\), \((3, 0)\), \((0, 3)\), and \((2, 0)\). To find the center of the circle passing through these points, we can find the midpoints of the segments formed by these points.

### Step 5: Find the midpoints of the segments.

**Midpoint of segment joining \((0, 2)\) and \((3, 0)\):**
\[ \text{Midpoint} = \left(\frac{0 + 3}{2}, \frac{2 + 0}{2}\right) = \left(\frac{3}{2}, 1\right) \]

**Midpoint of segment joining \((0, 3)\) and \((2, 0)\):**
\[ \text{Midpoint} = \left(\frac{0 + 2}{2}, \frac{3 + 0}{2}\right) = \left(1, \frac{3}{2}\right) \]

### Step 6: Find the center of the circle.

The center of the circle will be the intersection of the perpendicular bisectors of these segments. The coordinates of the midpoints are \(\left(\frac{3}{2}, 1\right)\) and \(\left(1, \frac{3}{2}\right)\).

### Step 7: Calculate the center.

Since the midpoints are symmetric about the line \(y = x\), the center of the circle can be found as:
\[ \text{Center} = \left(\frac{3/2 + 1}{2}, \frac{1 + 3/2}{2}\right) = \left(\frac{5/2}, \frac{5/2}\right) \]

### Final Answer:
The center of the circle is:
\[ \left(\frac{5}{2}, \frac{5}{2}\right) \]