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

##### 1 Answer

#### Answer:

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#

#### Explanation:

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

Let

The midpoint of

The slope of

So any perpendicular to

The perpendicular through the midpoint of

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

The points on this line are equidistant from

The midpoint of

The slope of

So any perpendicular to

The perpendicular through the midpoint of

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

The points on this line are equidistant from

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

#4 = -3x+5#

Subtract

#-1 = -3x#

Divide both sides by

#x = 1/3#

Then substitute this into the second equation to find

So the circle has equation

Substitute point

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

So

So our equation in standard form is:

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