A triangle has corners at #(5 ,1 )#, #(2 ,9 )#, and #(4 ,3 )#. What is the area of the triangle's circumscribed circle?

Question

A triangle has corners at #(5  ,1   )#, #(2   ,9   )#, and #(4   ,3  )#.  What is the area of the triangle's circumscribed circle?

Noah Atkinson · Accepted Answer

The area of the circle is #A = 1825/2pi#

When I do this type of problem, I always shift the 3 points so that one of them becomes the origin:

#(5,1) to (0,0)# #(2,9) to (-3,8)# #(4,3) to (-1, 2)#

The standard equation of a circle is:

#(x - h)^2 + (y - k)^2 = r^2" [1]"#

where #(x, y)# is any point on the circle, #(h, k)# is the center, and r is the radius.

Use equation [1] and the 3 shifted points to write 3 equations:

#(0 - h)^2 + (0 - k)^2 = r^2" [2]"# #(-3 - h)^2 + (8 - k)^2 = r^2" [3]"# #(-1 - h)^2 + (2 - k)^2 = r^2" [4]"#

Equation [2] simplifies to a very useful equation:

#h^2 + k^2 = r^2" [5]"#

Substitute the left side of equation [5] into the right sides of equations [3] and [4]:

#(-3 - h)^2 + (8 - k)^2 = h^2 + k^2" [6]"# #(-1 - h)^2 + (2 - k)^2 = h^2 + k^2" [7]"#

Expand the squares on the left side of equations [6] and [7]:

#9 + 6h + h^2 + 64 - 16k + k^2 = h^2 + k^2" [8]"# #1 + 2h + h^2 + 4 - 4k + k^2 = h^2 + k^2" [9]"#

#h^2 + k^2# is on both sides of the equations and this sums to zero:

#9 + 6h + 64 - 16k = 0" [10]"# #1 + 2h + 4 - 4k = 0" [11]"#

Collect the constant terms into a single term on the right:

#6h - 16k = -73" [12]"# #2h - 4k = -5" [13]"#

Multiply equation [13] by -4 and add to equation [12]:

#-2h = -53#

#h = 53/2#

Substitute #53/2# for h in equation [13};

#2(53/2) - 4k = -5#

#-4k = -58#

#k = 29/2#

Use equation [5] to compute #r^2#:

#r^2 = (53/2)^2 + (29/2)^2#

#r^2 = 2809/4 + 841/4#

#r^2 = 3650/4#

#r^2 = 1825/2#

The area of the circle is:

#A = pir^2#

#A = 1825/2pi#

Sophia Alfred · Answer

undefined To find the area of the circumscribed circle of a triangle, you can use the formula:

Area = (abc) / (4R)

Where a, b, and c are the lengths of the triangle's sides, and R is the radius of the circumscribed circle.

First, calculate the lengths of the sides using the distance formula:

a = √((5-2)^2 + (1-9)^2) = √(3^2 + (-8)^2) = √(9 + 64) = √73
b = √((2-4)^2 + (9-3)^2) = √((-2)^2 + (6)^2) = √(4 + 36) = √40
c = √((4-5)^2 + (3-1)^2) = √((-1)^2 + (2)^2) = √(1 + 4) = √5

Next, calculate the area:

Area = (√73 * √40 * √5) / (4R)

We need to find the radius (R) of the circumscribed circle. The radius of the circumscribed circle can be found using the formula:

R = (abc) / (4 * Area of the triangle)

Using Heron's formula to find the area of the triangle:

S = (a + b + c) / 2
Area of the triangle = √(S(S-a)(S-b)(S-c))

Where S is the semi-perimeter of the triangle.

S = ( √73 + √40 + √5 ) / 2 ≈ 10.24

Area of the triangle = √(10.24(10.24-√73)(10.24-√40)(10.24-√5))

Now that we have the area of the triangle, we can find R and then find the area of the circumscribed circle.

Finally, substitute the values of a, b, c, and the area of the triangle into the formula for R, and then use that value to calculate the area of the circumscribed circle using the formula provided initially.