# How do you find an equation of a circle passing through points (-6,-1) , (-4,3) , (2,-5)?

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Jan 22, 2016

${x}^{2} + {y}^{2} - 14 x - 10 y = 131$

#### Explanation:

Given 3 points

${P}_{1} \left(- 6. - 1\right) , {P}_{2} \left(- 4 , 3\right) , {P}_{3} \left(2 , - 5\right)$

For equation 1
${P}_{1} = {P}_{2}$

${\left(x + 6\right)}^{2} + {\left(y + 1\right)}^{2} = {\left(x + 4\right)}^{2} + {\left(y - 3\right)}^{2}$

By expanding, we have
${x}^{2} + 12 x + 36 + {y}^{2} + 2 y + 1 = {x}^{2} + 8 x + 16 + {y}^{2} - 6 y + 9$

By solving further, we get...

$4 x + 8 y = - 12$

$x + 2 y = - 3$ (Equation 1)

To solve for equation 2

We let

${P}_{2} = {P}_{3}$

${\left(x + 4\right)}^{2} + {\left(y - 3\right)}^{2} = {\left(x - 2\right)}^{2} + {\left(y + 5\right)}^{2}$

${x}^{2} + 8 x + 16 + {y}^{2} - 6 y + 9 = {x}^{2} - 4 x + 4 + {y}^{2} + 10 y + 25$

Using the same method we did on equation 1, we get...

$12 x - 16 y = 4$

$3 x - 4 y = 1$ (Equation 2)

We can find the center by solving the 2 equations obtained by elimination method.

$- 2 \left[x + 2 y = - 3\right]$
$3 x - 4 y = 1$

$- 2 x + 4 y = 6$
$3 x - 4 y = 1$

$x = 7$

$3 \left(7\right) - 4 y = 1$

$21 - 4 y = 1$
$- 4 y = - 20$

$y = 5$

We now have the center of the circle

$C \left(7 , 5\right)$

To solve for the radius, we use the distance formula and substitute the center and either of the three points

${r}^{2} = {\left(- 6 - 7\right)}^{2} + {\left(- 1 - 5\right)}^{2}$
${r}^{2} = {\left(- 13\right)}^{2} + {\left(- 6\right)}^{2}$
${r}^{2} = 169 + 36$

${r}^{2} = 205$

We now have the components of the circle.

${\left(x - 7\right)}^{2} + {\left(y - 5\right)}^{2} = 205$

By expanding the equation...

${x}^{2} - 14 x + 49 + {y}^{2} - 10 y + 25 = 205$

By combining like terms and solving, we get...
${x}^{2} + {y}^{2} - 14 x - 10 y = 131$

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