Question

Find the equation of the circle.
x^2+y^2+Dx+Ey+F=0
that passes through the points: (2,0)(-1,-3)(5,-3)

Answer 1

simply input the points to the equation. when you put (2,0) you got equation1. : (2)^2 +2D +F = 0 => 2D+F=-4
Then the (-1,-3) and (5,-3) so that you get 2 more equations :
equation 2: -D-3E+F=-10 and equation 3: 5D-3E+F =-34
then if you do elimination on equation 2 and 3, so that you'll get
-6D=24 . Therefore D= -4
then put the value of D on equation 1. you'll get F=-4+8 = 4.
then put the value of D and F to equation 2 or 3 and you'll get E=-18.
Therefore the equation of the circle is : x^2 + y^2 - 4x - 18y +4 = 0

Answer 2

i think i got the value of E wrong. the value of E should be 6 . then the right equation is ^2 + y^2 - 4x - 6y +4 = 0

Answer 3

ups sorry ,typo. the equation is x^2 + y^2 - 4x + 6y +4 = 0.

Answer 4

WOw this makes a lot more sense than my book! Thank you!

Answer 5

you're welcome :)