The circumcenter of a triangle is the point where the perpendicular bisectors of the sides intersect. This point is equidistant from all three vertices of the triangle.

Given the vertices $A(5,6)$, $B(-2,7)$, and $C(4,3)$, we will find the circumcenter by following these steps:

Step 1: Find the midpoints of the sides.

• Midpoint of $AB$ is: $M_{AB} = \left( \frac{5 + (-2)}{2}, \frac{6 + 7}{2} \right) = \left( \frac{3}{2}, \frac{13}{2} \right)$

• Midpoint of $BC$ is: $M_{BC} = \left( \frac{-2 + 4}{2}, \frac{7 + 3}{2} \right) = (1, 5)$

• Midpoint of $CA$ is: $M_{CA} = \left( \frac{5 + 4}{2}, \frac{6 + 3}{2} \right) = \left( \frac{9}{2}, \frac{9}{2} \right)$

Step 2: Find the slopes of the sides.

• Slope of $AB$: $\text{slope of } AB = \frac{7 - 6}{-2 - 5} = \frac{1}{-7} = -\frac{1}{7}$

• Slope of $BC$: $\text{slope of } BC = \frac{3 - 7}{4 - (-2)} = \frac{-4}{6} = -\frac{2}{3}$

• Slope of $CA$: $\text{slope of } CA = \frac{6 - 3}{5 - 4} = \frac{3}{1} = 3$

Step 3: Find the slopes of the perpendicular bisectors.

The slopes of the perpendicular bisectors are the negative reciprocals of the slopes of the sides:

• Slope of perpendicular bisector of $AB$: $\text{slope of perpendicular bisector of } AB = 7$

• Slope of perpendicular bisector of $BC$: $\text{slope of perpendicular bisector of } BC = \frac{3}{2}$

• Slope of perpendicular bisector of $CA$: $\text{slope of perpendicular bisector of } CA = -\frac{1}{3}$

Step 4: Write the equations of the perpendicular bisectors.

Using the point-slope form of the line equation $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is the midpoint:

1. Equation of the perpendicular bisector of $AB$ (slope = 7, point = $M_{AB}(1.5, 6.5)$): $y - 6.5 = 7(x - 1.5) \implies y = 7x - 10$

2. Equation of the perpendicular bisector of $BC$ (slope = $\frac{3}{2}$, point = $M_{BC}(1, 5)$): $y - 5 = \frac{3}{2}(x - 1) \implies y = \frac{3}{2}x + \frac{7}{2}$

Step 5: Find the intersection of the two perpendicular bisectors.

Set the equations equal to each other:

$7x - 10 = \frac{3}{2}x + \frac{7}{2}$

Multiply through by 2 to eliminate the fraction: $14x - 20 = 3x + 7$

Simplify and solve for $x$: $14x - 3x = 20 + 7 \implies 11x = 27 \implies x = \frac{27}{11}$

Substitute $x = \frac{27}{11}$ into one of the equations to find $y$. Using $y = 7x - 10$:

$y = 7 \times \frac{27}{11} - 10 = \frac{189}{11} - \frac{110}{11} = \frac{79}{11}$

So, the circumcenter is $\left( \frac{27}{11}, \frac{79}{11} \right)$.

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Would you like more detailed steps or explanations on any part of this? Or perhaps some related questions?

Related Questions:

1. How do you find the orthocenter of a triangle using the altitudes?
2. What is the relationship between the circumcenter and the centroid of a triangle?
3. How can the circumradius be calculated once you have the circumcenter?
4. What does the circumcenter represent in the context of a triangle's geometry?
5. How would you find the circumcenter if the triangle is right-angled?

Tip:

When working with perpendicular bisectors, it’s helpful to check that the slopes are correctly calculated, especially if using fractions, as any small mistake can lead to an incorrect intersection point.