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

Dec 11, 2016

$A r e a = \frac{145 \pi}{18}$

#### Explanation:

The equation of a circle is:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2} \text{ [1]}$

where, $\left(x , y\right)$ is any point on the circle, $\left(h , k\right)$ is the center point, and r is the radius.

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:

$\left(5 , 1\right) \to \left(0 , 0\right)$
$\left(2 , 4\right) \to \left(- 3 , 3\right)$
$\left(7 , 2\right) \to \left(2 , 1\right)$

We write the three equations using the new points:

${\left(0 - h\right)}^{2} + {\left(0 - k\right)}^{2} = {r}^{2} \text{ [2]}$
${\left(- 3 - h\right)}^{2} + {\left(3 - k\right)}^{2} = {r}^{2} \text{ [3]}$
${\left(2 - h\right)}^{2} + {\left(1 - k\right)}^{2} = {r}^{2} \text{ [4]}$

Equation [2] simplifies:

${h}^{2} + {k}^{2} = {r}^{2} \text{ [5]}$

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

${\left(- 3 - h\right)}^{2} + {\left(3 - k\right)}^{2} = {h}^{2} + {k}^{2} \text{ [6]}$
${\left(2 - h\right)}^{2} + {\left(1 - k\right)}^{2} = {h}^{2} + {k}^{2} \text{ [7]}$

Expand the squares:

$9 + 6 h + {h}^{2} + 9 - 6 k + {k}^{2} = {h}^{2} + {k}^{2} \text{ [8]}$
$4 - 4 h + {h}^{2} + 1 - 2 k + {k}^{2} = {h}^{2} + {k}^{2} \text{ [9]}$

The ${h}^{2} \mathmr{and} {k}^{2}$ terms cancel:

$9 + 6 h + 9 - 6 k = 0 \text{ [10]}$
$4 - 4 h + 1 - 2 k = 0 \text{ [11]}$

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

$6 h - 6 k = - 18 \text{ [10]}$
$- 4 h - 2 k = - 5 \text{ [11]}$

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

$18 h = - 3$

$h = - \frac{1}{6}$

Substitute $- \frac{1}{6}$ for h into equation [11]:

$- 4 \left(- \frac{1}{6}\right) - 2 k = - 5$

$- 2 k = - 5 - \frac{2}{3}$

$- 2 k = - \frac{17}{3}$

$k = \frac{17}{6}$

Use equation [2] to find the value of ${r}^{2}$:

${r}^{2} = {\left(- \frac{1}{6}\right)}^{2} + {\left(\frac{17}{6}\right)}^{2}$

${r}^{2} = \frac{290}{36} = \frac{145}{18}$

$A r e a = \pi {r}^{2}$

$A r e a = \frac{145 \pi}{18}$

Dec 11, 2016

≈25.31" square units"

#### Explanation:

The area (A) of the circle is $\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{A = \pi {r}^{2}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
To calculate the area we require the radius of the circle.

The $\textcolor{b l u e}{\text{circumcentre}}$ of the circle is at the intersection of the 3 $\textcolor{b l u e}{\text{perpendicular bisectors}}$ 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.
$\textcolor{m a \ge n t a}{\text{----------------------------------------------------------------}}$
$\textcolor{b l u e}{\text{Equations of perpendicular bisectors}}$

$\text{using the form } \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y - {y}_{1} = m \left(x - {x}_{1}\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{ and "color(blue)"gradient formula } m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

$\text{with } {m}_{p e r p} = - \frac{1}{m}$

$\text{side } \left(5 , 1\right) \to \left(2 , 4\right)$

$\text{mid-point } = \left[\frac{1}{2} \left(5 + 2\right) , \frac{1}{2} \left(1 + 4\right)\right] = \left(\frac{7}{2} , \frac{5}{2}\right)$

$m = \frac{4 - 1}{2 - 5} = \frac{3}{- 3} = - 1 \Rightarrow {m}_{p e r p} = 1$

$\Rightarrow y - \frac{5}{2} = x - \frac{7}{2} \Rightarrow y = x - 1 \to \left(1\right)$

$\text{side } \left(5 , 1\right) \to \left(7 , 2\right)$

$\text{mid-point } = \left[\frac{1}{2} \left(5 + 7\right) , \frac{1}{2} \left(1 + 2\right)\right] = \left(6 , \frac{3}{2}\right)$

$m = \frac{2 - 1}{7 - 5} = \frac{1}{2} \Rightarrow {m}_{p e r p} = - 2$

$\Rightarrow y - \frac{3}{2} = - 2 \left(x - 6\right) \Rightarrow y = - 2 x + \frac{27}{2} \to \left(2\right)$
$\textcolor{m a \ge n t a}{\text{-------------------------------------------------------------------}}$

$\textcolor{b l u e}{\text{Finding the circumcentre}}$

Using equations (1) and (2) from above, both expressed in terms of y.

$\Rightarrow x - 1 = - 2 x + \frac{27}{2} \Rightarrow 3 x = \frac{29}{2} \Rightarrow x = \frac{29}{6}$

substituting in (1): $y = \frac{29}{6} - 1 = \frac{23}{6}$

$\text{Hence circumcentre } = \left(\frac{29}{6} , \frac{23}{6}\right)$
$\textcolor{m a \ge n t a}{\text{--------------------------------------------------------------}}$

$\textcolor{b l u e}{\text{calculating the radius (r) and area}}$

$\text{r is the distance from } \left(\frac{29}{6} , \frac{23}{6}\right)$ to one of the 3 vertices.

$\text{Using " (29/6,23/6)to(2,4)"with the"color(blue)" distance formula}$

$r = \sqrt{{\left(\frac{29}{6} - 2\right)}^{2} + {\left(\frac{23}{6} - 4\right)}^{2}}$

$= \sqrt{{\left(\frac{17}{6}\right)}^{2} + {\left(- \frac{1}{6}\right)}^{2}}$

$\Rightarrow {r}^{2} = {\left(\frac{17}{6}\right)}^{2} + {\left(- \frac{1}{6}\right)}^{2} = \frac{289}{36} + \frac{1}{36} = \frac{290}{36}$

$\text{Thus area of circumcircle " =290/36pilarr" exact value}$

$\Rightarrow \text{area " ≈25.31" to 2 decimal places}$