Question

What is the equation of a circle that passes through point J(6, 4) and has center K (1, -8).

Answer 1

please help!!! :o

Answer 2

Use the distance formula to calculate the length of the radius.

Answer 3

um can you show me? like give me an example of how i would set it up and everything

Answer 4

i don't know :(

Answer 5

What is the distance formula?

Answer 6

D= sr(x2-x1)^2 + (y2-y1)^2

Answer 7

Now plug in: the two points are: (6,4) and (1,-8)

Answer 8

So x2=8
x1=6
y2=-8
y1=4

Answer 9

x= 8? wouldn't it be 1 though?

Answer 10

where did you get 2 8's at, ugh, this is confusing well i did 6-1 = 5 and 4 - 8 = -4 . ... so 5 - 4 = 1 , is 1 right?

Answer 11

Sorry. x2=1

Answer 12

oh wait i forget to square them

Answer 13

so it would be like 20 - 16?

Answer 14

which is 4 and the square root of 4 is 2 ... idk :(

Answer 15

\[d=\sqrt{(1-6)^2+(-8-4)^2}=\sqrt{(-5)^2+(-12)^2}=\]

Answer 16

Now you finish it.

Answer 17

169

Answer 18

13

Answer 19

thank you

Answer 20

Can you finish the problem now that you know the radius is 13?

Answer 21

yes thank you lots :o

Answer 22

yw