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

1 Answer
Jan 5, 2017

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

Explanation:

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:

  • For point (5,6)

(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)

  • For point (4,3)

(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)

  • For point (2,2)

(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:

  • Equations (1) minus (2) results in:

9-2x_0+27-6y_0=0

x_0=18-3y_0 Equation (4)

  • Equations (2) minus (3) results in:

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