Find The Coordinates of the Circumcenter...?

Question

With this information:

Triangle FGH with verticies F (0,-6), G (3,6), and H (12,0).

Can you please show me how to do this step by step?

Thanks in advance!!!

Answer 1

The circumcenter is the point in a triangle where you can put your compass and draw a circle (centered there) and have the circle touch each of the vertexes of the triangle. The circumcenter is found by drawing in the perpendicular bisectors of each of the sides. So here's what you need to do:

1) Find the equation of the perpendicular bisectors of two of the sides (that requires you to know the midpoint of the sides, as well as the slope).

2) Find the intersection of the two perpendicular bisectors.

Midpoint FG = ((0+3)/2, (-6+6)/2) = (3/2, 0)
Slope FG = (6--6)/(3-0) = 12/3 = 4

Equation of perpendicular bisector of FG:
y = -(1/4)x + B
0 = -(1/4)(3/2) + B
0 = -3/8 + B
B = 3/8
y = -(1/4)X + 3/8


Midpoint of GH = ((3+12)/2, (6+0)/2) = (15/2, 3)
Slope of GH = (0-6)/(12-3) = -6/9 = -2/3

Equation of perpendicular bisector of GH:
y = (3/2)x + B
3 = (3/2)(15/2) + B
3 = 45/4 + B
B = -33/4
y = (3/2)X - 33/4


Now we need the intersection of:
y = -(1/4)X + 3/8
y = (3/2)X - 33/4

-(1/4)X + 3/8 = (3/2)X - 33/4
x = 69/14
y = -6/7

Answer: (69/14, -6/7)

Answer 2

Just to recap: the circumcenter is the center of the circle that circumscribes the triangle. Since the three points are going to be on the circle, we need to find the point that's equidistant to all 3.

There are a couple of ways to go about this. One way is to use the distance formula: √((x2-x1)²+(y2-y1)²). If our unkown point is at (x,y) and this is equidistant to the other 3, then:
√((x-0)²+(y+6)²) = √((x-3)²+(y-6)²) = √((x-12)²+(y-0)²)

Take the first and third distances and set them equal to each other:
√((x)²+(y+6)²) = √((x-12)²+(y)²)
x²+(y+6)² = (x-12)²+y²
x²+(y²+12y+36) = (x²-24x+144)+y²
12y+36 = -24x+144
y + 3 = -2x + 12
y = -2x + 9

Now take a different pair, simplify:
√((x-0)²+(y+6)²) = √((x-3)²+(y-6)²)
x²+(y+6)² = (x-3)²+(y-6)²
x²+(y²+12y+36) = (x²-6x+9)+(y²-12y+36)
(x²+y²+36)+12y = (x²+y²+36) + (-6x+9)+(-12y)
12y = (-6x+9)+(-12y)
24y = -6x + 9
8y = -2x + 3

You've got two equations with two unknowns. Plug y = -2x + 9 into the second equation:
8(-2x + 9) = -2x + 3
-16x + 72 = -2x + 3
69 = 14x
x = 69/14

So y = -2(69/14) + 9 = -69/7 + 63/7 = -6/7

The coordinates are then (69/14, -6/7)

Answer 3

The circumcenter is the point of concurrence of the three perpendicular bisectors.

Step 1. Use the midpoint formula to find the middle of a side

Step 2. Use the point slope formula to find the equation of the line through the two points.

Step 3. Recall that a line perpendicular to a given line has a slope that is the negative reciprocal of the slope of the given line. and use the point slope formula again to find the equation of the perpendicular line through the midpoint

Step 4. Return to step 1 and do it again for one of the other sides.

Step 5. Finally, find the point of intersection of the two lines. That is the circumcenter.

Good luck.