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

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Answer 1 · 2015-12-29T18:19:58

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

$(x-1/3)^2+(y+5/3)^2 = ((10 sqrt(2))/3)^2$

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

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

The midpoint of $AB$ is $((-3+1)/2, (-5+3)/2) = (-1, -1)$

The slope of $AB$ is $(3-(-5))/(1-(-3)) = 8/4 = 2$

So any perpendicular to $AB$ has slope $-1/2$

The perpendicular through the midpoint of $AB$ can be written:

$color(blue)(y+1 = -1/2 (x+1))$

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

The midpoint of $BC$ is $((1+5)/2, (3-1)/2) = (3, 1)$

The slope of $BC$ is $(-1-3)/(5-1) = (-4)/4 = -1$

So any perpendicular to $BC$ has slope $-(1/-1) = 1$

The perpendicular through the midpoint of $BC$ can be written:

$color(blue)(y-1 = x-3)$

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

Subtract the two equations of the perpendicular lines to find:

$2 = -1/2(x+1) - (x-3) =-3/2x+5/2$

Multiply both sides by $2$ to get:

$4 = -3x+5$

Subtract $5$ from both sides:

$-1 = -3x$

Divide both sides by $-3$ :

$x = 1/3$

Then substitute this into the second equation to find $y = -5/3$

So the circle has equation $(x-1/3)^2+(y+5/3)^2 = r^2$ for some radius $r$ .

Substitute point $B$ into this equation to find:

$r^2 = (1-1/3)^2 + (3+5/3)^2 =4/9+ 196/9 = 200/9$

So $r = sqrt(200/9) = (10 sqrt(2))/3$

So our equation in standard form is:

$(x-1/3)^2+(y+5/3)^2 = ((10 sqrt(2))/3)^2$