A triangle has corners at #(5 ,6 )#, #(4 ,3 )#, and #(2 ,2 )#. What is the area of the triangle's circumscribed circle?

Question

A triangle has corners at #(5  ,6   )#, #(4   ,3   )#, and #(2   ,2  )#.  What is the area of the triangle's circumscribed circle?

Violet Aldrich · Accepted Answer

To solve use the following:

Area of circle
#A=π*r^2#
- Equation of circle
  #(x-x_0)^2+(y-y_0)^2=r^2#
  - Know that all corners are points of the circle, so they satisfy the
    equation
    Answer is:
    
    #A=π*12.5~=39.27# The area of the circle is: #A=π*r^2# Where #r# is the radius. To find the radius, we will use the equation of the circle: #(x-x_0)^2+(y-y_0)^2=r^2# Where #x_0# and #y_0# are the coordinates of the circles center. This equation holds for all of the circle's points, including the three corners of the triangle . Therefore, we have three equations: #(5-x_0)^2+(6-y_0)^2=r^2# By expanding the identities: #25-10x_0+(x_0)^2+36-12y_0+(y_0)^2=r^2# Equation (1) #(4-x_0)^2+(3-y_0)^2=r^2# By expanding the identities: #16-8x_0+(x_0)^2+9-6y_0+(y_0)^2=r^2# Equation (2) #(2-x_0)^2+(2-y_0)^2=r^2# By expanding the identities: #4-4x_0+(x_0)^2+4-4y_0+(y_0)^2=r^2# Equation (3) Now from these three equations we can find #x_0# and #y_0#. To do so, we substract the equations in two pairs. Here I will substract the following: #9-2x_0+27-6y_0=0# #x_0=18-3y_0# Equation (4) #12-4x_0+5-2y_0=0# Equation (5) Substitute equation (4) in (5): #12-4(18-3y_0)+5-2y_0=0# #y_0=5.5# Equation (6) Substitute equation (6) in equation (4) to find #x_0#: #x_0=18-3*5.5# #x_0=1.5# Now that the coordinates of the circles center are known, the equation of the circle can be taken for any of the three known points. Let's pick for example point (2,2): #(x-x_0)^2+(y-y_0)^2=r^2# #(x-1.5)^2+(y-5.5)^2=r^2# #(2-1.5)^2+(2-5.5)^2=r^2# #0.5^2+(-3.5)^2=r^2# #r^2=12.5# Finally, the area of the circle is: #A=π*r^2=π*12.5~=39.27#

Allison Allen · Answer

undefined To find the area of the circumscribed circle of a triangle, we need to first find the circumradius $ R $, which is the radius of the circle that passes through all three vertices of the triangle. Then, we can use the formula for the area of a circle, $ A = \pi R^2 $.

To find the circumradius $ R $, we can use the formula:

$$ R = \frac{abc}{4A} $$

Where $ a $, $ b $, and $ c $ are the lengths of the sides of the triangle, and $ A $ is the area of the triangle.

We can find the lengths of the sides of the triangle using the distance formula:

$$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

Once we have the lengths of the sides, we can use Heron's formula to find the area of the triangle:

$$ A = \sqrt{s(s-a)(s-b)(s-c)} $$

Where $ s $ is the semi-perimeter of the triangle, given by $ s = \frac{a + b + c}{2} $.

Finally, we can substitute the values of $ a $, $ b $, $ c $, and $ A $ into the formula for $ R $ and then use $ R $ to find the area of the circumscribed circle.