# Finding center and radius of circumscribed triangle

Let $u=(6,7), v=(-2, 3), w=(7,0)$. Find the center and the radius of the circumscribed circle of triangle $[u,v,w]$.

My approach would look like this:

(1) Determine the angles of the triangle $[u,v,w]$ via dot products.

(2) If we draw three lines from the the center $o$ of the circle to the three vertices, the triangle $[u,v,w]$ will be made of three equilateral triangles. We can then easily find from (1) all the nine angles of the three triangles.

(3) Determine the heights of the three inscribed triangles, which will be the same for all of the three, and which will be the radius $r$ of the inscribed circle.

(4) Then we can drop a perpendicular line from $o$ to, say, $[v,w]$ in order to find the coordinates of $o$.

But is there a much less tedious way to solve this problem? And do you think my approach is correct?

• With regard to your step (2): In general, you get isosceles triangles, not equilateral triangles. – quasi May 26 '17 at 5:46
• @quasi Thanks, that's what I was going to mean actually. – sequence May 26 '17 at 5:53

HINT:

If $(h,k)$ is the center with radius $=r$

$$r^2=(h-6)^2+(k-7)^2=(h+2)^2+(k-3)^2=(h-7)^2+(k-0)^2$$

$$(h-6)^2+(k-7)^2=(h+2)^2+(k-3)^2\ \ \ \ (1)$$

and $$(h+2)^2+(k-3)^2=(h-7)^2+(k-0)^2\ \ \ \ (2)$$

will give two simultaneous equations in $h,k$

• Have you looked up "eructation" in the dictionary? – Lord Shark the Unknown May 26 '17 at 5:47
• @LordSharktheUnknown, Not sure if I've understood u – lab bhattacharjee May 26 '17 at 5:49
• I certainly haven't understood you! – Lord Shark the Unknown May 26 '17 at 5:50
• @LordSharktheUnknown, Then why don't you pinpoint the are of confusion? – lab bhattacharjee May 26 '17 at 5:54
• "simultaneous eructations": an evocative, but not terribly complimentary metaphor! – Lord Shark the Unknown May 26 '17 at 5:58

Let $x^2+y^2+Dx+Ey+F=0$ be the equation of the circle. Then

$$\begin{cases} 6^2+7^2+6D+7E+F=0 \\ (-2)^2+3^2-2D+3E+F=0 \\ 7^2+7D+F=0 \end{cases}$$

$$\begin{cases} 6D+7E+F=-85 \\ -2D+3E+F=-13 \\ 7D+F=-49 \end{cases}$$

Solving, $D=-6$, $E=-6$ and $F=-7$.

The equation of the circle is

$$x^2+y^2-6x-6y-7=0$$

$$(x-3)^2+(y-3)^2=25$$

The centre is $(3,3)$ and the radius is $5$.

$(2,5)$ is a middle point of $UV$.

$m_{UV}=\frac{7-3}{6+2}=\frac{1}{2}$.

Thus, the center placed in the line $$y-5=-2(x-2)$$ or $$y=-2x+9.$$ Let $(x,-2x+9)$ is our center.

Thus, $$(x+2)^2+(-2x+6)^2=(x-7)^2+(-2x+9)^2,$$ which gives $(3,3)$ is the center and the rest is smooth: $$(x-3)^2+(y-3)^2=25$$

Let $(h,k)$ be the coordinates of the circumcenter, and let $r$ denote the circumradius.

The circumcenter is the point where the perpendicular bisectors meet.

You only need two of them, any two.

• The segment $uv$ has slope $\frac{1}{2}$ and midpoint $(2,5)$.
• The segment $vw$ has slope $-\frac{1}{3}$ and midpoint $(\frac{5}{2},\frac{3}{2})$

It follows that $h,k$ satisfy the slope equations \begin{align*} \frac{k-5}{h-2}=-2\\[6pt] \frac{k-\frac{3}{2}}{h-\frac{5}{2}}=3\\[6pt] \end{align*}

Thus, two linear equations, two unknowns.

Solving the system yields $(h,k)=(3,3)$.

To find $r$, just use the distance formula to get the distance from $(3,3)$ to any of the three vertices.

Thus, choosing the vertex $(7,0)$, we get \begin{align*} r&=\sqrt{(3 - 7)^2 + (3 - 0)^2}\\[4pt] &=\sqrt{25}\\[4pt] &=5 \end{align*}

• Can you please clarify how you get these slope equations? – sequence May 30 '17 at 3:34
• Since the line segment $uv$ has slope $1/2$, the perpendicular bisector of $uv$ has slope $-2$ (the negative reciprocal of $1/2$), so the slope of the line segment from the midpoint of $uv$ to $(h,k)$ is equal to $-2$. Similarly, since the line segment $vw$ has slope $-1/3$, the perpendicular bisector of $vw$ has slope $3$, so the slope of the line segment from the midpoint of $vw$ to $(h,k)$ is equal to $3$. – quasi May 30 '17 at 5:20

I have an unusual approach to this problem that I was excited to learn about when I first saw it.

The equation of a circle through points $u = (6,7)$, $v = (-2,3)$, and $w = (7,0)$ can be written down directly as the determinant equation $$\begin{vmatrix} x^2+y^2 & x & y & 1 \\ 6^2+7^2 & 6 & 7 & 1 \\ (-2)^2 + 3^2 & -2 & 3 & 1 \\ 7^2 + 0^2 & 7 & 0 & 1\end{vmatrix}=0$$ which we may simplify to $$60 x^2-360 x+60 y^2-360 y-420 = 0 \iff (x-3)^2 + (y-3)^2 = 25.$$ (In the last step, we divide by $60$, then complete the square twice.) From here, we can deduce that the circle has center $(3,3)$ and radius $5$.

We can prove that this method works in two simple steps:

1. First, any determinant equation like this expands to have the form $$A(x^2+y^2) + Bx + Cy + D = 0$$ and the graph of such an equation is usually a circle.
2. Second, if you plug in the coordinates of one of the given points $u$, $v$, or $w$, the determinant is $0$ (because the matrix we get has two identical rows), so the graph we get does pass through the points $u,v,w$.

The method fails gracefully on bad input: we get a line instead of a circle when the three points are collinear, and the determinant is $0$ for any $(x,y)$ when the three points are not all distinct.

Unlike all other methods, this one generalizes easily to finding a sphere through four points (see this page, which is where I first learned about it) and probably to many other similar problems.

• Can you please explain how exactly you come up with your determinant? – sequence May 30 '17 at 4:23
• We make one row be $(x^2+y^2, x, y, 1)$ and the rest constants so that the determinant is $0$ on some circle, because circles have equations of the form $A(x^2+y^2)+Bx+Cy+D=0$. We make the other rows be $(x_u^2+y_u^2, x_u, y_u, 1)$ and similar so that the determinant is $0$ when $x=x_u$ and $y=y_u$. – Misha Lavrov May 30 '17 at 10:49