Question

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

Answer 1

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`