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

The equation of a circle passing through the vertices of a triangle can be found using the **circumcircle** of the triangle. In this case, we need to find the circumcircle of the triangle formed by the line \( 2x + 3y = 6 \), the x-axis, and the y-axis.

### Step 1: Find the vertices of the triangle
The triangle's vertices are the points where the line intersects the x-axis and y-axis, along with the origin.

1. **Intersection with the x-axis:**  
   The x-axis is defined by \( y = 0 \). Substituting \( y = 0 \) into the line equation:
   \[
   2x + 3(0) = 6 \implies x = 3
   \]
   So, the point is \( (3, 0) \).

2. **Intersection with the y-axis:**  
   The y-axis is defined by \( x = 0 \). Substituting \( x = 0 \) into the line equation:
   \[
   2(0) + 3y = 6 \implies y = 2
   \]
   So, the point is \( (0, 2) \).

3. **The origin:**  
   The third vertex is the origin \( (0, 0) \), where the x-axis and y-axis meet.

Thus, the vertices of the triangle are \( (0, 0) \), \( (3, 0) \), and \( (0, 2) \).

### Step 2: Find the circumcenter
To find the equation of the circle, we need its center (the circumcenter) and its radius.

The circumcenter of a right triangle (which this is, since two sides lie along the axes) is the midpoint of the hypotenuse.

- The hypotenuse connects the points \( (3, 0) \) and \( (0, 2) \).

The midpoint of the hypotenuse is:
\[
\left( \frac{3 + 0}{2}, \frac{0 + 2}{2} \right) = \left( \frac{3}{2}, 1 \right)
\]
Thus, the circumcenter is \( \left( \frac{3}{2}, 1 \right) \).

### Step 3: Find the radius
The radius of the circumcircle is the distance from the circumcenter to any vertex of the triangle. Let's calculate the distance from \( \left( \frac{3}{2}, 1 \right) \) to \( (0, 0) \).

Using the distance formula:
\[
\text{Distance} = \sqrt{ \left( \frac{3}{2} - 0 \right)^2 + \left( 1 - 0 \right)^2 } = \sqrt{ \left( \frac{3}{2} \right)^2 + 1^2 } = \sqrt{ \frac{9}{4} + 1 } = \sqrt{ \frac{9}{4} + \frac{4}{4} } = \sqrt{ \frac{13}{4} } = \frac{\sqrt{13}}{2}
\]
Thus, the radius \( r = \frac{\sqrt{13}}{2} \).

### Step 4: Write the equation of the circle
The general form of the equation of a circle is:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
where \( (h, k) \) is the center and \( r \) is the radius.

Substitute \( h = \frac{3}{2} \), \( k = 1 \), and \( r^2 = \frac{13}{4} \):
\[
\left( x - \frac{3}{2} \right)^2 + (y - 1)^2 = \frac{13}{4}
\]

Multiplying the entire equation by 4 to eliminate the fraction:
\[
4 \left( x - \frac{3}{2} \right)^2 + 4(y - 1)^2 = 13
\]

Expanding this:
\[
4 \left( x^2 - 3x + \frac{9}{4} \right) + 4(y^2 - 2y + 1) = 13
\]
\[
4x^2 - 12x + 9 + 4y^2 - 8y + 4 = 13
\]
\[
4x^2 + 4y^2 - 12x - 8y + 13 = 13
\]
\[
4x^2 + 4y^2 - 12x - 8y = 0
\]

Thus, the equation of the circle is:
\[
4x^2 + 4y^2 - 12x - 8y = 0
\]

Would you like a more detailed explanation or have any questions?

### 5 related questions:
1. How do you find the circumcenter of a general triangle?
2. What are the properties of a right triangle’s circumcircle?
3. Can a triangle always have a circumcircle?
4. How do you find the radius of a circle given its equation?
5. What is the difference between the incenter and circumcenter of a triangle?

### Tip:
To quickly find the circumcenter of a right triangle, remember that it lies at 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.

Discover how to find the equation of the circle that passes through the vertices of the triangle formed by the line 2x + 3y = 6, the x-axis, and the y-axis. This detailed step-by-step solution covers key concepts in analytic geometry, including finding the circumcenter and radius using coordinate geometry. Suitable for students in Grades 10-12.

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