# How do you find the equation of the circle in general form containing the points (-3,-5), (1,3), and (5,-1)?

Dec 29, 2015

Use an alternative, slightly more geometric method (not necessarily shorter) to find:

${\left(x - \frac{1}{3}\right)}^{2} + {\left(y + \frac{5}{3}\right)}^{2} = {\left(\frac{10 \sqrt{2}}{3}\right)}^{2}$

#### Explanation:

The centre of the circle can also be found as the intersection of two lines using linear equations as follows:

Let $A = \left(- 3 , - 5\right)$, $B = \left(1 , 3\right)$ and $C = \left(5 , - 1\right)$

The midpoint of $A B$ is $\left(\frac{- 3 + 1}{2} , \frac{- 5 + 3}{2}\right) = \left(- 1 , - 1\right)$

The slope of $A B$ is $\frac{3 - \left(- 5\right)}{1 - \left(- 3\right)} = \frac{8}{4} = 2$

So any perpendicular to $A B$ has slope $- \frac{1}{2}$

The perpendicular through the midpoint of $A B$ can be written:

$\textcolor{b l u e}{y + 1 = - \frac{1}{2} \left(x + 1\right)}$

The points on this line are equidistant from $A$ and $B$.

The midpoint of $B C$ is $\left(\frac{1 + 5}{2} , \frac{3 - 1}{2}\right) = \left(3 , 1\right)$

The slope of $B C$ is $\frac{- 1 - 3}{5 - 1} = \frac{- 4}{4} = - 1$

So any perpendicular to $B C$ has slope $- \left(\frac{1}{-} 1\right) = 1$

The perpendicular through the midpoint of $B C$ can be written:

$\textcolor{b l u e}{y - 1 = x - 3}$

The points on this line are equidistant from $B$ and $C$.

Subtract the two equations of the perpendicular lines to find:

$2 = - \frac{1}{2} \left(x + 1\right) - \left(x - 3\right) = - \frac{3}{2} x + \frac{5}{2}$

Multiply both sides by $2$ to get:

$4 = - 3 x + 5$

Subtract $5$ from both sides:

$- 1 = - 3 x$

Divide both sides by $- 3$:

$x = \frac{1}{3}$

Then substitute this into the second equation to find $y = - \frac{5}{3}$

So the circle has equation ${\left(x - \frac{1}{3}\right)}^{2} + {\left(y + \frac{5}{3}\right)}^{2} = {r}^{2}$ for some radius $r$.

Substitute point $B$ into this equation to find:

${r}^{2} = {\left(1 - \frac{1}{3}\right)}^{2} + {\left(3 + \frac{5}{3}\right)}^{2} = \frac{4}{9} + \frac{196}{9} = \frac{200}{9}$

So $r = \sqrt{\frac{200}{9}} = \frac{10 \sqrt{2}}{3}$

So our equation in standard form is:

${\left(x - \frac{1}{3}\right)}^{2} + {\left(y + \frac{5}{3}\right)}^{2} = {\left(\frac{10 \sqrt{2}}{3}\right)}^{2}$