Question

how do i turn the end points of a circle (1,1) and (-1,-1) in general form equation of a circle?

Answer 1

end points of a circle? i assume you mean that is its diameter

Answer 2

endpoints of a diameter of a circle?

Answer 3

yes diameter

Answer 4

find the midpoint; in this case it (0,0) which makes life easier

Answer 5

yes

Answer 6

midpoint of a diameter of a circle is the center of the circle

Answer 7

and recall that the radius from (0,0) to (1,1) = sqrt(2)

Answer 8

using distance formula to find the radius :)

Answer 9

then fill in the circle equation: x^2 + y^2 = r^2

Answer 10

if r=sqrt{2}, then r^2=?

Answer 11

i think general form tho is what: x^2 + y^2 - r^2?

Answer 12

\[(x-x _{1})(x-x _{2})+(y-y _{1})(y-y _{1})=0\]

Answer 13

y-y2 in the last part

Answer 14

x^2+y^2=r^2 is cool
for centers not (0,0) use
(x-h)^2+(y-k)^2=r^2

Answer 15

Substitute the values and you will eventually get the equation

Answer 16

where (h,k) is center

Answer 17

for centers not zero i throw my hands in the air and get some hagen daas

Answer 18

how can u substitue 0s and get general form :/?

Answer 19

(x-0)^2 = x^2

Answer 20

same concept for the y part

Answer 21

(h,k)=(0,0)
then you have
(x-0)^2+(y-0)^2=r^2
same as
x^2+y^2=r^2

Answer 22

ok later guys
i have something to take care of

Answer 23

to find the midpoint; i find it easier to add my points and divide it by 2

( 1 , 1)
(-1,-1)
-------
( 0 , 0 ) /2 = (0,0)

Answer 24

same

Answer 25

so then i also plug in 1,1 as x and y right?

Answer 26

i call my center point: C(xy) to which

(x-Cx)^2 + (y-Cy)^2 - r^2 = 0 would be the general form

Answer 27

no; x and y are umm.. dummy variables that stand in for whatever it is we need them for

Answer 28

Cx and Cy are what you should fill in

Answer 29

oh
so X^2+Y^2=R^2

Answer 30

this is the gen form ?

Answer 31

well, i answered prematurely; that is the standard form i believe

Answer 32

ok

Answer 33

well its simplified so isnt it gen ?

Answer 34

http://www.mathsisfun.com/algebra/circle-equations.html

Answer 35

terminology gets me half the time; but standard from resembles the pythag thrm; general form is long and drawn out

Answer 36

in this case its rather short and sweet;

x^2 + y^2 -2 = 0

Answer 37

wow had got this answer but it looked short reason i got confused

Answer 38

thats becasue its centered at 0 and the extra stuff disappears

Answer 39

thanks

Answer 40

youre welcome :)