HELP!! give an equation for a circle that goes through (1, -3), (-3, 1) and (-5, -1)

We could solve this with some Algebra....but ...it's messy...a Geometric solution is actually easier

(1, -3), (-3, 1) and (-5, -1).

Find the midpoint  and slope between the first two points

Midpoint [ (1 - 3)/ 2 , (-3 + 1)/2)] =  (-2/2, -2/2) = (-1, -1)

Slope   =  [ 1 - - 3] / [ -3 - 1] = [ 4/ -4]  = -1

And the negative reciprocal of the slope = 1

So.....using the midpoint and the negative reciprocal slope, the equation of the line with this slope and passing through the midpoint is  y = 1 (x - - 1) - 1    ⇒  y = x + 1  - 1   ⇒    y = x        (1)  

Now using any of the other two points [ I'll use (1, -3) and (-5, - 1) ]  we want to do he same thing

Midpoint  [ ( -5 + 1) / 2 , (-3 -1 ) / 2 ]  = [ ( -4/ 2 ), ( -4/2) ] = (-2, -2)

Slope between these points =  [ - 1 - - 3] / [ -5 - 1 ] =  [ 2/-6] = (-1/3)

And the negative reciprocal slope is  3

So...using this slope and the midpoint....we can write the equation of the line as

y = 3(x - - 2) -2 ⇒   y = 3x + 6  - 2 ⇒   y = 3x  + 4     (2)

Find the x coordinate of the intersection of  (1) and (2)....this will be he x coordinate of the center of the circle

x = 3x + 4

-2x = 4

x = - 2

And  using y = x

The y coordinate of this intersection is -2

So  the center of this circle is (-2 , - 2)

Snd the distance from this center to any of the points is the radius....so ...using the point (1, -3)....the radius^2 is

(-2 - 1)^2  + ( -2 - -3)^2  =

(-3)^2 + (1)^2 =

10

So.....the equation of the circle passing through these 3 points is

(x + 2)^2 + (y + 2)^2  = 10

Here's the graph: