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. A. (2, 3)
B. B. (3, 2)
C. C. (5, 5)
D. 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 intersect the axes. The equations given are:

1. \(2x + 3y - 6 = 0\)
2. \(9x + 6y - 18 = 0\)

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

To find the x-intercept, set \(y = 0\):
\[ 2x + 3(0) - 6 = 0 \implies 2x - 6 = 0 \implies 2x = 6 \implies x = 3 \]
Thus, the x-intercept is \((3, 0)\).

To find the y-intercept, set \(x = 0\):
\[ 2(0) + 3y - 6 = 0 \implies 3y - 6 = 0 \implies 3y = 6 \implies y = 2 \]
Thus, the y-intercept is \((0, 2)\).

The points where the first line cuts the axes are \((3, 0)\) and \((0, 2)\).

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

To find the x-intercept, set \(y = 0\):
\[ 9x + 6(0) - 18 = 0 \implies 9x - 18 = 0 \implies 9x = 18 \implies x = 2 \]
Thus, the x-intercept is \((2, 0)\).

To find the y-intercept, set \(x = 0\):
\[ 9(0) + 6y - 18 = 0 \implies 6y - 18 = 0 \implies 6y = 18 \implies y = 3 \]
Thus, the y-intercept is \((0, 3)\).

The points where the second line cuts the axes are \((2, 0)\) and \((0, 3)\).

### Step 3: Identify the four points.

The four points we have are:
1. \(A(3, 0)\)
2. \(B(0, 2)\)
3. \(C(2, 0)\)
4. \(D(0, 3)\)

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

For four points to be concyclic, the opposite angles must sum to \(180^\circ\) or we can use the determinant method. However, we can also find the center of the circle that passes through these points.

### Step 5: Find the center of the circle.

The center of the circle that passes through the points can be found by calculating the midpoints of the segments connecting the points.

1. Midpoint of \(AB\):
\[ M_{AB} = \left(\frac{3 + 0}{2}, \frac{0 + 2}{2}\right) = \left(\frac{3}{2}, 1\right) \]

2. Midpoint of \(CD\):
\[ M_{CD} = \left(\frac{2 + 0}{2}, \frac{0 + 3}{2}\right) = \left(1, \frac{3}{2}\right) \]

### Step 6: Find the intersection of the perpendicular bisectors.

The slope of line \(AB\) is \(-\frac{2}{3}\), so the slope of the perpendicular bisector is \(\frac{3}{2}\). The equation of the perpendicular bisector through \(M_{AB}\) is:
\[ y - 1 = \frac{3}{2}\left(x - \frac{3}{2}\right) \]

The slope of line \(CD\) is \(-\frac{3}{2}\), so the slope of the perpendicular bisector is \(\frac{2}{3}\). The equation of the perpendicular bisector through \(M_{CD}\) is:
\[ y - \frac{3}{2} = \frac{2}{3}(x - 1) \]

### Step 7: Solve these two equations to find the center.

1. From the first equation:
\[ y = \frac{3}{2}x - \frac{9}{4} + 1 = \frac{3}{2}x - \frac{5}{4} \]

2. From the second equation:
\[ y = \frac{2}{3}x + \frac{3}{2} - \frac{2}{3} = \frac{2}{3}x + \frac{7}{6} \]

Set them equal to find \(x\):
\[ \frac{3}{2}x - \frac{5}{4} = \frac{2}{3}x + \frac{7}{6} \]

Solving this will give us the coordinates of the center.

### Final Result:

After solving the equations, we find that the center of the circle is at \(\left(\frac{5}{2}, \frac{5}{2}\right)\).