# A triangle has corners at #(5 ,1 )#, #(2 ,4 )#, and #(7 ,2 )#. What is the area of the triangle's circumscribed circle?

##### 2 Answers

#### Explanation:

The equation of a circle is:

where,

Before we use equation [1] and the 3 given points to write 3 equations, let's move the points so that one of them is the origin. This will not affect the area of the circumscribed circle; it only moves the circle to a different location:

We write the three equations using the new points:

Equation [2] simplifies:

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

Expand the squares:

The

collect the constant terms into an single term on the right:

Multiply equation [11] by -3 and add to equation [10]:

Substitute

Use equation [2] to find the value of

#### Explanation:

The area (A) of the circle is

#color(red)(bar(ul(|color(white)(2/2)color(black)(A=pir^2)color(white)(2/2)|)))#

To calculate the area we require the radius of the circle.The

#color(blue)"circumcentre"# of the circle is at the intersection of the 3#color(blue)"perpendicular bisectors"# of the sides of the triangle. The perpendicular bisector, bisects the side of a triangle at right angles.To find the centre we only require 2 equations of perpendicular bisectors, then solve to find coordinates of the centre.

The distance then from the circumcentre to any of the 3 vertices will provide us with the radius and thus area.

#color(magenta)"----------------------------------------------------------------"#

#color(blue)"Equations of perpendicular bisectors"#

#"using the form " color(red)(bar(ul(|color(white)(2/2)color(black)(y-y_1=m(x-x_1))color(white)(2/2)|)))#

#" and "color(blue)"gradient formula "m=(y_2-y_1)/(x_2-x_1)#

#"with " m_(perp)=-1/m#

#"side " (5,1)to(2,4)#

#"mid-point " =[1/2(5+2),1/2(1+4)]=(7/2,5/2)#

#m=(4-1)/(2-5)=3/(-3)=-1rArrm_(perp)=1#

#rArry-5/2=x-7/2rArry=x-1to(1)#

#"side " (5,1)to(7,2)#

#"mid-point "=[1/2(5+7),1/2(1+2)]=(6,3/2)#

#m=(2-1)/(7-5)=1/2rArrm_(perp)=-2#

#rArry-3/2=-2(x-6)rArry=-2x+27/2to(2)#

#color(magenta)"-------------------------------------------------------------------"#

#color(blue)"Finding the circumcentre"# Using equations (1) and (2) from above, both expressed in terms of y.

#rArrx-1=-2x+27/2rArr3x=29/2rArrx=29/6# substituting in (1):

#y=29/6-1=23/6#

#"Hence circumcentre " =(29/6,23/6)#

#color(magenta)"--------------------------------------------------------------"#

#color(blue)"calculating the radius (r) and area"#

#"r is the distance from " (29/6,23/6)# to one of the 3 vertices.

#"Using " (29/6,23/6)to(2,4)"with the"color(blue)" distance formula"#

#r=sqrt((29/6-2)^2+(23/6-4)^2)#

#=sqrt((17/6)^2+(-1/6)^2)#

#rArrr^2=(17/6)^2+(-1/6)^2=289/36+1/36=290/36#

#"Thus area of circumcircle " =290/36pilarr" exact value"#

#rArr"area " ≈25.31" to 2 decimal places"#