Naudiapie
  • Naudiapie
What is the equation of a circle that passes through point J(6, 4) and has center K (1, -8).
Mathematics
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SOLVED
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katieb
  • katieb
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Naudiapie
  • Naudiapie
please help!!! :o
Mertsj
  • Mertsj
Use the distance formula to calculate the length of the radius.
Naudiapie
  • Naudiapie
um can you show me? like give me an example of how i would set it up and everything

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Naudiapie
  • Naudiapie
i don't know :(
Mertsj
  • Mertsj
What is the distance formula?
Naudiapie
  • Naudiapie
D= sr(x2-x1)^2 + (y2-y1)^2
Mertsj
  • Mertsj
Now plug in: the two points are: (6,4) and (1,-8)
Mertsj
  • Mertsj
So x2=8 x1=6 y2=-8 y1=4
Naudiapie
  • Naudiapie
x= 8? wouldn't it be 1 though?
Naudiapie
  • Naudiapie
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?
Mertsj
  • Mertsj
Sorry. x2=1
Naudiapie
  • Naudiapie
oh wait i forget to square them
Naudiapie
  • Naudiapie
so it would be like 20 - 16?
Naudiapie
  • Naudiapie
which is 4 and the square root of 4 is 2 ... idk :(
Mertsj
  • Mertsj
|dw:1437270786863:dw|
Mertsj
  • Mertsj
\[d=\sqrt{(1-6)^2+(-8-4)^2}=\sqrt{(-5)^2+(-12)^2}=\]
Mertsj
  • Mertsj
Now you finish it.
Naudiapie
  • Naudiapie
169
Naudiapie
  • Naudiapie
13
Naudiapie
  • Naudiapie
thank you
Mertsj
  • Mertsj
Can you finish the problem now that you know the radius is 13?
Naudiapie
  • Naudiapie
yes thank you lots :o
Mertsj
  • Mertsj
yw

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