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.