The equation of the circumcircle of the triangle formed by the lines x=0, y=0, 2x+3y=5, is

Options

A. `6(x^(2)+y^(2))+5(3x-2y)=0`
B. `x^(2)+y^(2)+2x-3y+5=0`
C. `x^(2)+y^(2)+2x-3y-5=0`
D. `6(x^(2)+y^(2))-5(3x+2y)=0`

Answer

To find the equation of the circumcircle of the triangle formed by the lines \(x = 0\), \(y = 0\), and \(2x + 3y = 5\), we can follow these steps:

### Step 1: Identify the Points of Intersection
1. **Point A**: Intersection of \(x = 0\) and \(2x + 3y = 5\):
- Substitute \(x = 0\) into the equation \(2x + 3y = 5\):
\[ 2(0) + 3y = 5 \implies 3y = 5 \implies y = \frac{5}{3} \]
- Thus, Point A is \((0, \frac{5}{3})\).

2. **Point B**: Intersection of \(y = 0\) and \(2x + 3y = 5\):
- Substitute \(y = 0\) into the equation \(2x + 3y = 5\):
\[ 2x + 3(0) = 5 \implies 2x = 5 \implies x = \frac{5}{2} \]
- Thus, Point B is \((\frac{5}{2}, 0)\).

3. **Point O**: The origin, which is \((0, 0)\).

### Step 2: Find the Midpoint of AB
- The midpoint \(M\) of line segment \(AB\) can be calculated using the midpoint formula:
\[ M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]
- Where \(A(0, \frac{5}{3})\) and \(B(\frac{5}{2}, 0)\):
\[ M = \left(\frac{0 + \frac{5}{2}}{2}, \frac{\frac{5}{3} + 0}{2}\right) = \left(\frac{5/2}{2}, \frac{5/3}{2}\right) = \left(\frac{5}{4}, \frac{5}{6}\right) \]

### Step 3: Calculate the Radius
- The radius \(r\) of the circumcircle is the distance from the center \(M\) to any vertex of the triangle, say point \(O(0, 0)\):
\[ r = \sqrt{\left(\frac{5}{4} - 0\right)^2 + \left(\frac{5}{6} - 0\right)^2} \]
\[ = \sqrt{\left(\frac{5}{4}\right)^2 + \left(\frac{5}{6}\right)^2} = \sqrt{\frac{25}{16} + \frac{25}{36}} \]
- To add these fractions, find a common denominator (which is 144):
\[ = \sqrt{\frac{25 \cdot 9}{144} + \frac{25 \cdot 4}{144}} = \sqrt{\frac{225 + 100}{144}} = \sqrt{\frac{325}{144}} = \frac{\sqrt{325}}{12} \]

### Step 4: Write the Equation of the Circle
- The standard form of the equation of a circle with center \((h, k)\) and radius \(r\) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
- Here, \(h = \frac{5}{4}\), \(k = \frac{5}{6}\), and \(r^2 = \frac{325}{144}\):
\[ \left(x - \frac{5}{4}\right)^2 + \left(y - \frac{5}{6}\right)^2 = \frac{325}{144} \]

### Step 5: Simplify the Equation
- Expanding the left side:
\[ \left(x - \frac{5}{4}\right)^2 = x^2 - \frac{5}{2}x + \frac{25}{16} \]
\[ \left(y - \frac{5}{6}\right)^2 = y^2 - \frac{5}{3}y + \frac{25}{36} \]
- Combining these:
\[ x^2 + y^2 - \frac{5}{2}x - \frac{5}{3}y + \left(\frac{25}{16} + \frac{25}{36}\right) = \frac{325}{144} \]
- Now, find a common denominator for \(\frac{25}{16}\) and \(\frac{25}{36}\):
\[ \frac{25 \cdot 9}{144} + \frac{25 \cdot 4}{144} = \frac{225 + 100}{144} = \frac{325}{144} \]
- Thus, the equation simplifies to:
\[ x^2 + y^2 - \frac{5}{2}x - \frac{5}{3}y = 0 \]

### Final Equation
- Multiplying through by 144 to eliminate fractions:
\[ 144x^2 + 144y^2 - 360x - 240y = 0 \]
- Dividing by 12 gives:
\[ 12x^2 + 12y^2 - 30x - 20y = 0 \]

### Conclusion
The equation of the circumcircle of the triangle formed by the lines is:
\[ 6x^2 + y^2 - 5(3x + 2y) = 0 \]