# Question #7a456

Oct 9, 2014

A circle is in the form

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

where $\left(h , k\right)$ is the center and r is the radius

Substituting known points in the circle, we have

[1] ${\left(2 - h\right)}^{2} + {\left(8 - k\right)}^{2} = {r}^{2}$
[2] ${\left(- 4 - h\right)}^{2} + {\left(6 - k\right)}^{2} = {r}^{2}$
[3] ${\left(1 - h\right)}^{2} + {\left(- 1 - k\right)}^{2} = {r}^{2}$

Since they are all equal to ${r}^{2}$, we can interchange the equations
[1] = [2]
${\left(2 - h\right)}^{2} + {\left(8 - k\right)}^{2} = {\left(- 4 - h\right)}^{2} + {\left(6 - k\right)}^{2}$
$\implies \left(4 - 4 h + {h}^{2}\right) + \left(64 - 16 k + {k}^{2}\right) = \left(16 - 8 h + {h}^{2}\right) + \left(36 - 12 k + {k}^{2}\right)$

[1] = [3]
${\left(2 - h\right)}^{2} + {\left(8 - k\right)}^{2} = {\left(1 - h\right)}^{2} + {\left(- 1 - k\right)}^{2}$
$\implies \left(4 - 4 h + {h}^{2}\right) + \left(64 - 16 k + {k}^{2}\right) = \left(1 - 2 h + {h}^{2}\right) + \left(1 + 2 k + {k}^{2}\right)$

Move all to the left side and simplify

[1] = [2]
$\implies \left(4 - 4 h\right) + \left(64 - 16 k\right) - \left(16 - 8 h\right) - \left(36 - 12 k\right) = 0$
$\implies 16 + 8 h - 4 k = 0$
$\implies 8 + 4 h - 2 k = 0$

[1] = [3]
$\implies \left(4 - 4 h\right) + \left(64 - 16 k\right) - \left(1 - 2 h\right) - \left(1 + 2 k\right) = 0$
$\implies 66 - 2 h - 18 k = 0$

[1'] $8 + 4 h - 2 k = 0$
[2'] $66 - 2 h - 18 k = 0$

Eliminate one of the variables by scaling any of the equations
such that the variable will have the same coefficient (different sign is OK) as with the other equation

[1'] $8 + 4 h - 2 k = 0$
[2'] $132 - 4 h - 18 k = 0$

Add (or subtract if different sign)

$140 - 20 k = 0$
$140 = 20 k$
$k = 7$

[1'] $8 + 4 h - 2 \left(7\right) = 0$
$4 h = 6$
$h = \frac{3}{2}$

the circle is centered at $\left(\frac{3}{2} , 7\right)$