Circumscribed triangle

Using the general form of a circle: we have ${ x }^{ 2 }+{ y }^{ 2 }+ax+by+c=0$ .

Substituting in the values of x and y for each of the three co-ordinates:

$3a+5b+c=-34\\ 2a-2b+c=-8\\ -6a+2b+c=-40$

Equation 1 - Equation 2:

$a+7b=-26$ $\rightarrow$ Equation 4

Equation 2 - Equation 3:

$8a-4b=32$ $\rightarrow$ Equation 5

Equation 4 * 8: $8a+56b=-208$ $\rightarrow$ Equation 6

Equation 6 - Equation 5:

$60b=-240\\ b=-4$

Substituting $b=-4$ into equation 4:

$a-28=-26\\ a=2$

Substituting $b=-4$ and $a=2$ into equation 2:

$12+c=-8\\c=-20$

So we have ${ x }^{ 2 }+{ y }^{ 2 }+2x-4y-20=0$ .

= ${ (x+1) }^{ 2 }+(y-2)^{ 2 }=25$

h=-1

k=2

r=5

h+k+r= $\boxed { 6 }$

Who knew a question from Kumon could become Level 3 here!!

From the equation of a circle: $(x-h)^2 + (y-k)^2 = r^2$ $(h,k)$ is the coordinate of the center of the circle, $r$ is the radius and $(x,y)$ is any point that lies in the circle.

Given that the circle circumscribes the triangle means that the three points lie in the circle which will satisfy the following three equations:

$(3-h)^2 + (5-k)^2 = r^2 \\ (2-h)^2 + (-2-k)^2 = r^2 \\ (-6-h)^2 + (2-k)^2 = r^2$

To solve for $h$ and $k$ we have to subtract the first equation from the following two which will give us two equations: $(2-h)^2 - (3-h)^2 + (-2-k)^2 - (5-k)^2 = r^2 - r^2 \\ \Rightarrow{(h^2 - 4h + 4) - (h^2 - 6h + 9) \\ + (k^2 + 4k + 4) - (k^2 - 10k + 25) = 0} \\ \Rightarrow{(2h -5) + (14k -21) = 0} \\ \Rightarrow{2h + 14k - 26 = 0} \rightarrow \text{(equation 1)} \\$

$(-6-h)^2 - (3-h)^2 + (2-k)^2 - (5-k)^2 = r^2 - r^2 \\ \Rightarrow{(h^2 + 12h + 36) - (h^2 - 6h + 9) \\ + (k^2 - 4k + 4) - (k^2 - 10k + 25) = 0} \\ \Rightarrow{(18h + 27) + (6k -21) = 0} \\ \Rightarrow{18h + 6k + 6 = 0} \rightarrow \text{(equation 2)} \\$

Solve the two equations $2h + 14k - 26 = 0 \\ 18h + 6k + 6 = 0$ we get: $h = \boxed{-1}$ and $k = \boxed{2}$ .

Finally to solve for $r$ you can get the first of the three equations then substitute $h$ and $k$ . $(3-h)^2 + (5-k)^2 = r^2 \\ \Rightarrow{(3-(-1))^2 + (5-(2))^2 = r^2 } \\ \Rightarrow{r = \sqrt{4^2 + 3^2}} \\ \Rightarrow{r = \sqrt{25}} \\ \Rightarrow{r = \boxed{5}} \\$

Therefor $h+k+r = \boxed{6}$