Write down the equation of the circle that passes through the vertices of the
triangle created by line 2x + 3y = 6, x-axis and y-axis.

To find the equation of the circle that passes through the vertices of the triangle created by the line \( 2x + 3y = 6 \), the x-axis, and the y-axis, we first need to determine the vertices of this triangle.

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

1. **Intersection with the x-axis**:  
   At the x-axis, \( y = 0 \).  
   Substituting into the equation of the line \( 2x + 3(0) = 6 \), we get:  
   \[
   2x = 6 \implies x = 3
   \]
   So, the line intersects the x-axis at \( (3, 0) \).

2. **Intersection with the y-axis**:  
   At the y-axis, \( x = 0 \).  
   Substituting into the equation of the line \( 2(0) + 3y = 6 \), we get:  
   \[
   3y = 6 \implies y = 2
   \]
   So, the line intersects the y-axis at \( (0, 2) \).

### Step 2: Find the vertices of the triangle.
The vertices of the triangle are:
- \( A(0, 0) \) (origin, the intersection of the axes),
- \( B(3, 0) \) (intersection with the x-axis),
- \( C(0, 2) \) (intersection with the y-axis).

### Step 3: Equation of the circumcircle.
The equation of the circle passing through the points \( A(0, 0) \), \( B(3, 0) \), and \( C(0, 2) \) can be written as:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
where \( (h, k) \) is the center of the circle and \( r \) is the radius.

Since the circumcenter of a right triangle (like this one) is located at the midpoint of the hypotenuse, we first need to find the midpoint of the hypotenuse, which is the line segment connecting points \( B(3, 0) \) and \( C(0, 2) \).

#### Midpoint of the hypotenuse:
The midpoint \( M \) of \( BC \) is:
\[
M = \left( \frac{3+0}{2}, \frac{0+2}{2} \right) = \left( \frac{3}{2}, 1 \right)
\]

So, the center of the circle is \( \left( \frac{3}{2}, 1 \right) \).

#### Step 4: Find the radius.
The radius of the circumcircle is the distance from the center \( M \left( \frac{3}{2}, 1 \right) \) to any of the vertices (say, \( A(0, 0) \)).

The distance from \( M \left( \frac{3}{2}, 1 \right) \) to \( A(0, 0) \) is given by the distance formula:
\[
r = \sqrt{ \left( \frac{3}{2} - 0 \right)^2 + (1 - 0)^2 } = \sqrt{ \left( \frac{3}{2} \right)^2 + 1^2 } = \sqrt{ \frac{9}{4} + 1 } = \sqrt{ \frac{13}{4} } = \frac{\sqrt{13}}{2}
\]

Thus, the radius is \( \frac{\sqrt{13}}{2} \).

### Step 5: Write the equation of the circle.
The equation of the circle is:
\[
\left( x - \frac{3}{2} \right)^2 + (y - 1)^2 = \left( \frac{\sqrt{13}}{2} \right)^2
\]
Simplifying:
\[
\left( x - \frac{3}{2} \right)^2 + (y - 1)^2 = \frac{13}{4}
\]
Multiplying through by 4 to eliminate the fraction:
\[
\left( 2x - 3 \right)^2 + 4(y - 1)^2 = 13
\]

This is the equation of the circumcircle passing through the vertices of the triangle.

Would you like any further details or clarifications on this process?

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### Related questions:
1. How can we find the equation of a circle given three arbitrary points?
2. What is the formula for the distance between two points in 2D space?
3. How do we derive the equation of a line from two points?
4. What are the properties of circumcenters in right triangles?
5. How can we generalize this approach for non-right triangles?

**Tip:** The circumcenter of a right triangle always lies on the midpoint of the hypotenuse.

Write down the equation of the circle that passes through the vertices of the triangle created by line 2x + 3y = 6, x-axis and y-axis.

Learn how to find the equation of a circle passing through the vertices of a triangle formed by the line 2x + 3y = 6, the x-axis, and the y-axis. This problem includes the use of the midpoint and distance formulas and explores the properties of circumcenters in right triangles.

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