Let C be the circle whose diameter is the line segment formed by the line `3x +2y = 6` intercepted by the coordinate axes, then C also passes through the point.

Options

A. `(2,2)`
B. `(1,2)`
C. `(2,3)`
D. `(3,2)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Find the intercepts of the line \(3x + 2y = 6\)

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

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

### Step 2: Determine the endpoints of the diameter

The endpoints of the diameter of the circle are the x-intercept and y-intercept:
- Point A: \((2, 0)\)
- Point B: \((0, 3)\)

### Step 3: Find the midpoint of the diameter

The center of the circle (midpoint) can be found using the midpoint formula:
\[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
Substituting the coordinates of points A and B:
\[ \left( \frac{2 + 0}{2}, \frac{0 + 3}{2} \right) = \left( 1, \frac{3}{2} \right) \]
So, the center of the circle is \((1, \frac{3}{2})\).

### Step 4: Calculate the radius of the circle

The radius is half the distance between the two endpoints of the diameter. We can use the distance formula to find the distance \(AB\):
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(0 - 2)^2 + (3 - 0)^2} = \sqrt{(-2)^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \]
Thus, the radius \(r\) is:
\[ r = \frac{AB}{2} = \frac{\sqrt{13}}{2} \]

### Step 5: Write the equation of the circle

The standard form of the equation of a circle is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \(h = 1\), \(k = \frac{3}{2}\), and \(r = \frac{\sqrt{13}}{2}\):
\[ (x - 1)^2 + \left(y - \frac{3}{2}\right)^2 = \left(\frac{\sqrt{13}}{2}\right)^2 \]
This simplifies to:
\[ (x - 1)^2 + \left(y - \frac{3}{2}\right)^2 = \frac{13}{4} \]

### Step 6: Check which points lie on the circle

We need to check which points satisfy the equation of the circle. Let's check the given points one by one.

1. **Point (2, 2)**:
\[ (2 - 1)^2 + \left(2 - \frac{3}{2}\right)^2 = 1^2 + \left(\frac{1}{2}\right)^2 = 1 + \frac{1}{4} = \frac{5}{4} \quad \text{(not equal to } \frac{13}{4}\text{)} \]

2. **Point (1, 2)**:
\[ (1 - 1)^2 + \left(2 - \frac{3}{2}\right)^2 = 0^2 + \left(\frac{1}{2}\right)^2 = 0 + \frac{1}{4} = \frac{1}{4} \quad \text{(not equal to } \frac{13}{4}\text{)} \]

3. **Point (2, 3)**:
\[ (2 - 1)^2 + \left(3 - \frac{3}{2}\right)^2 = 1^2 + \left(\frac{3}{2}\right)^2 = 1 + \frac{9}{4} = \frac{13}{4} \quad \text{(equal to } \frac{13}{4}\text{)} \]

4. **Point (4, 3)**:
\[ (4 - 1)^2 + \left(3 - \frac{3}{2}\right)^2 = 3^2 + \left(\frac{3}{2}\right)^2 = 9 + \frac{9}{4} = \frac{36}{4} + \frac{9}{4} = \frac{45}{4} \quad \text{(not equal to } \frac{13}{4}\text{)} \]

### Conclusion

The circle passes through the point **(2, 3)**.

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