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: Find the points of intersection
1. The line \( x = 0 \) intersects \( y = 0 \) at the origin \( O(0, 0) \).
2. The line \( x = 0 \) intersects \( 2x + 3y = 5 \):
- Substitute \( x = 0 \) into the equation:
\[ 2(0) + 3y = 5 \implies 3y = 5 \implies y = \frac{5}{3} \]
- This gives the point \( A(0, \frac{5}{3}) \).
3. The line \( y = 0 \) intersects \( 2x + 3y = 5 \):
- Substitute \( y = 0 \) into the equation:
\[ 2x + 3(0) = 5 \implies 2x = 5 \implies x = \frac{5}{2} \]
- This gives the point \( B(\frac{5}{2}, 0) \).

### Step 2: Identify the vertices of the triangle
The vertices of the triangle formed by the lines are:
- \( O(0, 0) \)
- \( A(0, \frac{5}{3}) \)
- \( B(\frac{5}{2}, 0) \)

### Step 3: Determine the circumcircle
Since \( O \) is the right angle of triangle \( OAB \), the hypotenuse \( AB \) will be the diameter of the circumcircle.

### Step 4: Find the midpoint of \( AB \)
The midpoint \( M \) of segment \( AB \) can be calculated as:
\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{0 + \frac{5}{2}}{2}, \frac{\frac{5}{3} + 0}{2} \right) = \left( \frac{5}{4}, \frac{5}{6} \right) \]

### Step 5: Calculate the radius
The radius \( r \) is half the distance between points \( A \) and \( B \):
\[ AB = \sqrt{ \left( \frac{5}{2} - 0 \right)^2 + \left( 0 - \frac{5}{3} \right)^2 } = \sqrt{ \left( \frac{5}{2} \right)^2 + \left( -\frac{5}{3} \right)^2 } = \sqrt{ \frac{25}{4} + \frac{25}{9} } \]
Finding a common denominator (36):
\[ AB = \sqrt{ \frac{225}{36} + \frac{100}{36} } = \sqrt{ \frac{325}{36}} = \frac{\sqrt{325}}{6} \]
Thus, the radius \( r = \frac{AB}{2} = \frac{\sqrt{325}}{12} \).

### Step 6: Write the equation of the circumcircle
The general equation of a circle is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Where \( (h, k) \) is the center and \( r \) is the radius. Substituting \( M \) and \( r \):
\[ \left( x - \frac{5}{4} \right)^2 + \left( y - \frac{5}{6} \right)^2 = \left( \frac{\sqrt{325}}{12} \right)^2 \]
This simplifies to:
\[ \left( x - \frac{5}{4} \right)^2 + \left( y - \frac{5}{6} \right)^2 = \frac{325}{144} \]

### Final Equation
The equation of the circumcircle of the triangle formed by the lines is:
\[ \left( x - \frac{5}{4} \right)^2 + \left( y - \frac{5}{6} \right)^2 = \frac{325}{144} \]