anonymous
  • anonymous
Which of the following circles have their centers in the second quadrant? Check all that apply. A. (x + 2)2 + (y - 5)2 = 9 B. (x - 4)2 + (y + 3)2 = 32 C. (x - 5)2 + (y - 6)2 = 25 D. (x + 1)2 + (y - 7)2 = 16
Mathematics
jamiebookeater
  • jamiebookeater
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butterflydreamer
  • butterflydreamer
first, can you tell me what the centre is for each of them?
butterflydreamer
  • butterflydreamer
To find the centre, use : centre of circle is at (h,k) equation:(x−h)2+(y−k)2=r2 So for e.g... for the first one A. \[(x+2)^2 + (y - 5)^2 = 9 \rightarrow centre (-2, 5) \]
anonymous
  • anonymous
wait why did the 2 turn negative

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butterflydreamer
  • butterflydreamer
because our general equation is this: \[(x-h)^2 + (y -k)^2 = r^2 , centre (h, k)\]
anonymous
  • anonymous
b- 4,-3
butterflydreamer
  • butterflydreamer
So when we compare it to: \[(x+2)^2 + (y - 5)^2 = 9, centre (-2, 5) \]
anonymous
  • anonymous
c- (5,6)
anonymous
  • anonymous
d- -1,7
anonymous
  • anonymous
is that right?
butterflydreamer
  • butterflydreamer
OKaaayyy let's seee :) A. centre (-2, 5) B. centre (4, -3) C. centre (5, 6) D. centre (-1, 7)
butterflydreamer
  • butterflydreamer
excellent work :)!
anonymous
  • anonymous
thank you :)
anonymous
  • anonymous
d and a are the answers
anonymous
  • anonymous
they lie in the second quadrant
butterflydreamer
  • butterflydreamer
no problem :) That's correct :D Wonderful!
anonymous
  • anonymous
thank you for your help!
butterflydreamer
  • butterflydreamer
you're welcome ^_^!
anonymous
  • anonymous
also,
anonymous
  • anonymous
The circle below is centered at the point (-2, 1), and has a radius of length 3. What is its equation? http://media.apexlearning.com/Images/200707/01/7f8d2e18-9709-467b-9bc7-1dc4df8eb1a9.gif A. (x + 2)2 + (y - 1)2 = 9 B. (x - 2)2 + (y + 1)2 = 3 C. (x + 1)2 + (y - 2)2 = 9 D. (x - 1)2 + (y + 2)2 = 3
anonymous
  • anonymous
when it asks for whats its equation, does it mean to fully work out the equation or just set it up?
butterflydreamer
  • butterflydreamer
Remember that the STANDARD equation of a circle is: \[(x-h)^2 + (y-k)^2 = r^2 \] where centre is at (h, k) So since they've told us that centre is at (-2, 1) , you want to find the equation. They've also told you that radius (r) =3 So now we know h = -2, k = 1 , r = 3 so sub this into the standard equation of a circle :)
anonymous
  • anonymous
it would be (x+2)^2(x-1)^2=3^2
anonymous
  • anonymous
choice a
butterflydreamer
  • butterflydreamer
yess :D That's correct
anonymous
  • anonymous
thank you :)
butterflydreamer
  • butterflydreamer
you're welcome :D!

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