# A triangle has corners at (3 ,7 ), (2 ,9 ), and (8 ,4 ). What is the area of the triangle's circumscribed circle?

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Mar 8, 2018

$\approx 166.2$ (see below)

#### Explanation:

If you just need the area of the circumscribed circle, the amount of calculations can be reduced a bit by appealing to a simple consequence of the sine law of triangles.

We usually express this law in the form

$\frac{a}{\sin} A = \frac{b}{\sin} B = \frac{c}{\sin} C$

what is often not mentioned is that this common value is, actually the diameter $d$ of the circumscribed circle. For a proof, see the figure below

Here ABC is a triangle whose circumscribed circle is drawn in red. AD is a diameter of the circle. Then $\angle C = \angle B D A$ (angles in the same segment of a circle) and $\Delta A B D$ is right-angled at B (angle in a semicircle). So

$\sin \angle C = \sin \angle A D B = \frac{A B}{A D} = \frac{c}{d} \implies d = \frac{c}{\sin \angle C}$

Armed with this result, we first calculate the lengths of the three sides of the given triangle. Let us label the vertices $\left(3 , 7\right)$, $\left(2 , 9\right)$, and $\left(8 , 4\right)$ by A, B, and C, respectively, Then

${a}^{2} = {\left(2 - 8\right)}^{2} + {\left(9 - 4\right)}^{2} = 36 + 25 = 61 \implies a = \sqrt{61}$
${b}^{2} = {\left(3 - 8\right)}^{2} + {\left(7 - 4\right)}^{2} = 25 + 9 = 34 \quad \implies b = \sqrt{34}$
${c}^{2} = {\left(3 - 2\right)}^{2} + {\left(7 - 9\right)}^{2} = 1 + 4 = 5 q \quad \quad \implies c = \sqrt{5}$

We also need one of the angles, say $\angle A$. For this we appeal to the cosine law for triangles :

${a}^{2} = {b}^{2} + {c}^{2} - 2 b c \cos A \implies \cos A = \frac{{b}^{2} + {c}^{2} - {a}^{2}}{2 b c}$

With our numbers, we have

$\cos A = \frac{34 + 5 - 61}{2 \sqrt{34} \sqrt{5}} = - \frac{11}{\sqrt{170}}$

Finally

${d}^{2} = {a}^{2} / \left\{{\sin}^{2} A\right\} = {a}^{2} / \left(1 - {\cos}^{2} A\right) = \frac{61}{1 - \frac{121}{170}} = \frac{61}{49} \times 170$

So, the area of the circumscribed circle is

$A = \pi {d}^{2} / 4 = \frac{\pi}{4} \times \frac{61}{49} \times 170 \approx 166.2$

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Tony B Share
Mar 8, 2018

area $\approx 166.216$ to 3 decimal places

$\textcolor{b r o w n}{\text{The approach by}}$ https://socratic.org/users/ananda-d $\textcolor{b r o w n}{\text{is so much simpler!}}$

#### Explanation:

Set point 1 as ${P}_{1} \to \left({x}_{1} , {y}_{1}\right) = \left(2 , 9\right)$
Set point 2 as ${P}_{2} \to \left({x}_{2} , {y}_{2}\right) = \left(3 , 7\right)$
Set point 3 as ${P}_{3} \to \left({x}_{3} , {y}_{3}\right) = \left(8 , 4\right)$

The perpendicular from $\frac{{P}_{1} + {P}_{3}}{2}$ will coincide with the perpendicular from $\frac{{P}_{2} + {P}_{3}}{2}$ at the centre of the circle
(circumcentre).

$\textcolor{b l u e}{\text{Consider mid point } {P}_{1} \to {P}_{3} = {P}_{4}}$

Set mid point as ${P}_{4}$

${x}_{4} = \frac{2 + 8}{2} = 5$
${y}_{4} = \frac{9 + 4}{2} = \frac{13}{2}$

${P}_{4} \to \left({x}_{4} , {y}_{4}\right) = \left(5 , \frac{13}{2}\right)$

$\textcolor{b l u e}{\text{Equation of line perpendicular to } {P}_{1} \to {P}_{3}}$

Set gradient ${P}_{1} \to {P}_{3} = m = \frac{{y}_{3} - {y}_{1}}{{x}_{3} - {x}_{1}} = \frac{4 - 9}{8 - 2} = - \frac{5}{6}$

Perpendicular gradient $= - \frac{1}{m} = + \frac{6}{5}$

So we have $\frac{6}{5} = \frac{y - \frac{13}{2}}{x - 5}$

$6 \left(x - 5\right) = 5 \left(y - \frac{13}{2}\right)$

$6 x - 30 = 5 y - \frac{65}{2}$

$6 x - 30 + \frac{65}{2} = 5 y$

$y = \frac{6}{5} x - \frac{30}{5} + \frac{65}{5 \times 2}$

$y = \frac{6}{5} x + \frac{1}{2} \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(1\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{\text{Consider mid point } {P}_{2} \to {P}_{3} = {P}_{5}}$

${x}_{5} = \frac{3 + 8}{2} = \frac{11}{2}$

${y}_{5} = \frac{7 + 4}{2} = \frac{11}{2}$

${P}_{5} \to \left({x}_{5} , {y}_{5}\right) = \left(\frac{11}{2} , \frac{11}{2}\right)$

Set gradient ${P}_{2} \to {P}_{3} = m = \frac{{y}_{3} - {y}_{2}}{{x}_{3} - {x}_{2}} = \frac{4 - 7}{8 - 3} = - \frac{3}{5}$

Perpendicular gradient$= - \frac{1}{m} = + \frac{5}{3}$

So we have $\frac{5}{3} = \frac{y - \frac{11}{2}}{x - \frac{11}{2}}$

$5 x - \frac{55}{2} = 3 y - \frac{33}{2}$

$5 x - 11 = 3 y$

$y = \frac{5}{3} x - \frac{11}{3} \text{ } \ldots . . E q u a t i o n \left(2\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine the centre of the circle}}$

Using simultaneous equations for $E q n \left(1\right) \mathmr{and} E q n \left(2\right)$ we get:

the centre of the circle is at: ${P}_{c} \to \left({x}_{c} , {y}_{c}\right) = \left(\frac{625}{70} , \frac{785}{70}\right)$

From this we can derive the radius and hence the area of the circle.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine the area of the circle}}$

Using Pythagoras on the two points ${P}_{c} \text{ and say } {P}_{3}$

Let the radius be $r$

$r = \sqrt{{\left({x}_{c} - {x}_{3}\right)}^{2} + {\left({y}_{c} - {y}_{3}\right)}^{2}}$

$r = \sqrt{{\left(\frac{625}{70} - 8\right)}^{2} + {\left(\frac{785}{70} - 4\right)}^{2}}$

$r = \sqrt{52.6904 \ldots}$

$r \approx 7.25881 \ldots \ldots$

area $= \pi {r}^{2} = 166.21589 \ldots$

area $\approx 166.216$ to 3 decimal places

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