Question

how do you find the center and radius of a circle with points (0,1) and (2,3).

Answer 1

those points are on the circle

Answer 2

(y-h)^2+(x-b)^2=r^2
r is radius
(h,b) is center of circle

Answer 3

but how can you use that if you don't already know the center or the radius

Answer 4

?

Answer 5

lol im thinking ofeasiest way to do it

Answer 6

haha okay, thank you

Answer 7

1st equation:
(1-h)^2+(0-b)^2=r^2
(1-h)^2+b^2=r^2
2nd equation:
(3-h)^2+(2-b)^2=r^2
combine equations:
(1-h)^2+b^2=(3-h)^2+(2-b)^2

Answer 8

okay, thanks so much!!

Answer 9

u still have to solve it tho lol

Answer 10

haha yes, i know. but that helps a lot knowing how to find center

Answer 11

after u solve that u should get:
h+b=3
or some variant

Answer 12

are (0,1) and (2,3) just two points on the circle?

Answer 13

or is there something else we know

Answer 14

yes

Answer 15

you have only two point. you are going to need something else.

Answer 16

not enough info i believe

Answer 17

then the problem can't be solved

Answer 18

all we know is that (0,1) and (2,3) are on the line. and we need to find center and radius.
this is in my pre-calc textbook as a homework question

Answer 19

i go to bed now
i have to get up at 5:00 am

Answer 20

line?

Answer 21

good night

Answer 22

oh a line. maybe we can do it then

Answer 23

is it a diameter?

Answer 24

myininaya! congratulations again. and you are so colorful!

Answer 25

thanks :)

Answer 26

you guys can do this one maybe if you guess correctly what he wants lol
i will leave you

Answer 27

@jahtoday post the entire problem and maybe we can solve it. two point and the line they are on is not enough

Answer 28

its on a circle

Answer 29

its not my post lol ive just been trying to solve it and i couldnt get any further

Answer 30

The problem is to find the center and radius of the circle. Write the standard form of the equation. The graph has a circle with two points labeled and those two points are (2,3) and (0,1). The center is not labeled

Answer 31

does the line go through the center?

Answer 32

If you guys aren't busy can you help me with my question?
Thanks :)

Answer 33

there is no line, its a circle. however the two points are on the diameter of the line

Answer 34

you mean diameter of the circle?

Answer 35

yes, diameter of the circle.

Answer 36

then the problem is simple

Answer 37

*smacks head against wall*!

Answer 38

lol

Answer 39

i just graphed it, and on my graph, because (0,1) and (2,3) was diameter, my graph shows (1,2) as center. is that correct?

Answer 40

yes

Answer 41

sorry, i should've explained more at the beginning :/

Answer 42

and then i can plug into the equation to find the radius?

Answer 43

what is the distance from the center to either of the points...

Answer 44

i haven't figured that out yet, do i need to?

Answer 45

yes

Answer 46

what will i use that for?

Answer 47

\[D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\]

Answer 48

but how will that help me find my radius?

Answer 49

the radius of a circle is the distance from the center of the circle to any point on the circle

Answer 50

oh so you can just divide by two?

Answer 51

no...you need to use the distance formula

Answer 52

once you use the distance formula, you divide by two right?

Answer 53

\[r=\sqrt{(2-1)^2+(3-2)^2}=\sqrt{1^2+1^2}=\sqrt{1+1}=\sqrt{2}\]

Answer 54

i also just plugged one of my coordinates and the center into the standard form and solved for radius, i believe it ends up being +/- square root of 2

Answer 55

we ended up with same thing!
thank you so much :)

Answer 56

that is because the formula for the circle is really the same as the distance formula

Answer 57

but i have one more question!! in the general equation x^2+y^2+Ax+By+C=0 where do A, B, and C come from

Answer 58

that is if you multiply out

\[(x-h)^2+(y-k)^2=r^2\]

Answer 59

oh it was the diameter! good work

Answer 60

yeah...took us a while but we figured that out ;)

Answer 61

in the general equation x^2+y^2+Ax+By+C=0 where do A, B, and C come from??

Answer 62

\[(x-h)^2+(y-k)^2=r^2\]

\[x^2-2xh+h^2+y^2-2yk+k^2-r^2=0\]

\[x^2+y^2+(-2h)x+(-2k)y+(h^2+k^2-r^2)=0\]

\[A=-2h\]

\[B=-2k\]

\[C=h^2+k^2-r^2\]

Answer 63

so you foiled out first. then completed the square?

Answer 64

like how would you write (x-1/2)^2+(y)^2=1/4

Answer 65

you don't need to complete the square

Answer 66

\[(x-1/2)^2+(y)^2=1/4\]

\[x^2-2\frac{1}{2}x+\frac{1}{4}+y^2-\frac{1}{4}=0\]

\[x^2+y^2+\left(-2\frac{1}{2}\right)x+0y+0=0\]

Answer 67

\[x^2+y^2-x=0\]