# Find the equation of the circle that passes through three points.

• Jan 5th 2010, 07:35 PM
Pre-Calc
Find the equation of the circle that passes through three points.
A ( 3, 1) B ( 2,4) C (-1, 2)

some things that go with this equation

-midpoint
-distance
-perpendicular bisector
-M1-M2= -1
-Simultaneous equations

this is what i have so far i don't know if its correct
(3-a)2 + (1-b)2 = r2 those 2's r squared im new to this and i just plugged in for the rest.

Correct me if im wrong please and help me out if you can
thank you.
• Jan 5th 2010, 10:22 PM
bigwave
Quote:

Originally Posted by Pre-Calc
A ( 3, 1) B ( 2,4) C (-1, 2)

some things that go with this equation

-midpoint
-distance
-perpendicular bisector
-M1-M2= -1
-Simultaneous equations

this is what i have so far i don't know if its correct
(3-a)2 + (1-b)2 = r2 those 2's r squared im new to this and i just plugged in for the rest.

Correct me if im wrong please and help me out if you can
thank you.

you are doing fine...

the standard equation of a circle is

$\left(x-h\right)^2 +\left(y-k\right)^2 = r^2$

from your equation a is h and b is K, it would probably be better to use the standard equation since that is what is commonly used.

first find the center of the circle by doing a simultaneous equations of the 2 perpendicular bisectors
of the lines AB and BC
then you have the center C(h,k)

to find the perpendicular bisector of line AB

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

squaring both sides yields
$\left(x-3\right)^2+\left(y-1\right)^2=\left(x-2\right)^2+\left(y-4\right)^2
$

next expand
$x^2-6x+9+y^2-2y+1=x^2-4x+4+y^2-8y+16$

$x-3y=-5$

continue with BC then find H,K by simultanuous equation
• Jan 6th 2010, 12:25 PM
Pre-Calc
Thank you I understand it now.