anonymous
  • anonymous
Give the center radius of each equation.Then sketch the graph. (x-3)square + (y+4)square minus 9=0
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
\[(x-3)^2+(y+4)^2=9\] if center = (h,k) and radius = r, then the general form is \[(x-h)^2+(y-k)^2=r^2\] Thus center=(3,-4) and r=3
anonymous
  • anonymous
why is 4 negative and 3 positive and thank so much
anonymous
  • anonymous
because it must be in the form (x-h)^2 and so for (x-h) to equal (x-3), h must be positive 3. Similarly, if (y+4)=(y-k), then k=-4

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anonymous
  • anonymous
ok how is this supose to be graphed because im looking at the example of the graph and it very difficult
anonymous
  • anonymous
do you have a compass?
anonymous
  • anonymous
no
anonymous
  • anonymous
well, it's going to be a circle, so I would, starting from (3,-4), go 3 units up, down, left and right.
anonymous
  • anonymous
then draw a curve through those four points by hand and make it as circular as you can?
anonymous
  • anonymous
ok one more question same problem different equation
anonymous
  • anonymous
x square + y square = 36
anonymous
  • anonymous
ok, this one is actually a little simpler. See, to get it in the standard form, all we do is rewrite it as \[(x-0)^2+(y-0)^2=6^2\] now clearly it's the same equation as the one you gave me, just a little more convenient
anonymous
  • anonymous
and from this we determine the center is (0,0) and the radius is 6
anonymous
  • anonymous
why is it 6 when you square root it
anonymous
  • anonymous
sorry dumb question
anonymous
  • anonymous
i see lol
anonymous
  • anonymous
that's alright, as long as you understand it now
anonymous
  • anonymous
and if you don't, i'd be happy to answer anything else
anonymous
  • anonymous
man you have been a big help ill be sure to be back on this mabey in a hour or so or minutes thanks :)

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