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.