# How do you write the equation for a circle that passes through the points at (4,5), (-2,3) and (-4,-3)?

Jun 26, 2016

${x}^{2} + {y}^{2} - 6 x + 4 y - 37 = 0$

#### Explanation:

let's substitute each coordinate of all points in the general equation of the circle:

${x}^{2} + {y}^{2} + a x + b y + c = 0$

so that you have three equations with unknown variables a,b,c:

${4}^{2} + {5}^{2} + 4 a + 5 b + c = 0$ and
${\left(- 2\right)}^{2} + {3}^{2} - 2 a + 3 b + c$ and
${\left(- 4\right)}^{2} + {\left(- 3\right)}^{2} - 4 a - 3 b + c = 0$

let's solve in a few steps:
4a+5b+c+41=0 and -2a+3b+c+13=0 and -4a-3b+c+25=0

$c = - 4 a - 5 b - 41 \mathmr{and}$(by substituting c) $- 2 a + 3 b - 4 a - 5 b - 41 + 13 = 0 \mathmr{and} - 4 a - 3 b - 4 a - 5 b - 41 + 25 = 0$

$- 6 a - 2 b - 28 = 0 \mathmr{and} - 8 a - 8 b - 16 = 0$

that is equivalent to:

$3 a + b + 14 = 0 \mathmr{and} a + b + 2 = 0$

By subtracting you obtain:

$2 a + 12 = 0$

$a = - 6$

so, by substituting the obtained value of a:

$- 6 + b + 2 = 0$

$b = 4$

so

$c = - 4 \left(- 6\right) - 5 \left(4\right) - 41 = 24 - 20 - 41 = - 37$

The equation of the circle is:

${x}^{2} + {y}^{2} - 6 x + 4 y - 37 = 0$