Find the circumcentre and circum radius of a triangle whose vertices are (3,0), (-1,-6) and (4,-1).?

Answer 1

The circumcenter (you'll have to put up with my American spellings) of a triangle is the center of a circle that contains the vertices, or has a circumradius that is equidistant to each of the vertices.

[mb5's intersections with the midpoints yield the centroid (center of gravity), not the circumcenter.]

Let (h, k) be your center, figure out the distance from the center to each point, then set them equal. The distance is the circumradius.

r² = (h - 3)² + k² = (h + 1)² + (k + 6)² = (h - 4)² + (k + 1)²

Using (h - 3)² + k² = (h + 1)² + (k + 6)²,
h² - 6h + 9 + k² = h² + 2h + 1 + k² + 12k + 36,
-8h - 28 = 12k, and
k = (-2h - 7) / 3.

Find k another way to enable a system of equations. (You have two variables, h and k, so you'll need two equations to solve for them.)

Using (h - 4)² + (k + 1)² = (h - 3)² + k²,
h² - 8h + 16 + k² + 2k + 1 = h² - 6h + 9 + k²,
2k = 2h - 8, and
k = h - 4.

You now have k = (-2h - 7)/3 = h - 4.

Solve for h, then solve for k and you'll have the center... the radius shouldn't be too tricky after that.

Good luck!

Answer 2

D = sqrt((x2 - x1)^2 + (y2 - y1)^2)

(3,0) and (-1,-6)
D = sqrt((3 - (-1))^2 + (0 - (-6))^2)
D = sqrt((3 + 1)^2 + (6)^2)
D = sqrt(4^2 + 36)
D = sqrt(16 + 36)
D = sqrt(52)
D = 2sqrt(13)

(-1,-6) and (4,-1)
D = sqrt((4 - (-1))^2 + (-6 - (-1))^2)
D = sqrt((4 + 1)^2 + (-5)^2)
D = sqrt(5^2 + 25)
D = sqrt(25 + 25)
D = sqrt(50)
D = 5sqrt(2)

(3,0) and (4,-1)
D = sqrt((4 - 3)^2 + (0 - (-1))^2)
D = sqrt(1^2 + 1^2)
D = sqrt(1 + 1)
D = sqrt(2)

So this is a scalene triangle

K = sqrt[s(s-a)(s-b)(s-c)]

s = (a + b + c)/2
s = (2sqrt(13) + 5sqrt(2) + sqrt(2))/2
s = (2sqrt(13) + (5 + 1)sqrt(2))/2
s = (2sqrt(13) + 6sqrt(2))/2
s = sqrt(13) + 3sqrt(2)

K = sqrt((sqrt(13) + 3sqrt(2))(sqrt(13) + 3sqrt(2) - 2sqrt(13))(sqrt(13) + 3sqrt(2) - 5sqrt(2))(sqrt(13) + 3sqrt(2) - sqrt(2)))

K = sqrt((sqrt(13) + 3sqrt(2))(3sqrt(2) - sqrt(13))(sqrt(13) - 2sqrt(2))(sqrt(13) + 2sqrt(2))

K = sqrt((sqrt(13) + sqrt(18))(sqrt(18) - sqrt(13))(sqrt(13) - sqrt(8))(sqrt(13) + sqrt(8))

K = sqrt((sqrt(18) + sqrt(13))(sqrt(18) - sqrt(13))(13 + sqrt(234) - sqrt(234) - 8))

K = sqrt((18 - sqrt(234) + sqrt(234) - 13)(5))
K = sqrt(5 * 5)
K = 5

Circumradius
R = abc/(4K)
R = (2sqrt(13) * 5sqrt(2) * sqrt(2))/(4 * 5)
R = ((2 * 5 * 1)sqrt(13 * 2 * 2))/20
R = (10sqrt(13 * 4))/20
R = (20sqrt(13))/20
R = sqrt(13) or about 3.61

Circumcentre
r = K/s
r = 5/(sqrt(13) + 3sqrt(2))
Multiply top and bottom by sqrt(13) - 3sqrt(2) to get rid of the sqrt
r = (5(sqrt(13) - 3sqrt(2)))/((sqrt(13) - 3sqrt(2))(sqrt(13) + 3sqrt(2))
r = (5(sqrt(13) - 3sqrt(2)))/(13 - 18)
r = (5(sqrt(13) - 3sqrt(2)))/(-5)
r = -(sqrt(13) - 3sqrt(2)) or about .6371

ANS :
R = sqrt(13) or about 3.61
r = -(sqrt(13) - 3sqrt(2)) or about .6371

I got all the work from the sites listed below. Sorry if i did it the longer way around in comparison to what others probably did.

Answer 3

You should not try to use Yahoo! Answers to do your homework. But if you want a hint on how to solve this type of problem, notice that line segments drawn from each vertex to the midpoint of the opposite edge will all intersect at the same point, which can be used as the center of a circle....

The rest is up to you.

Answer 4

Graph it and use a compass