Question

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

Answer 1

first, can you tell me what the centre is for each of them?

Answer 2

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.
\[(x+2)^2 + (y - 5)^2 = 9 \rightarrow centre (-2, 5) \]

Answer 3

wait why did the 2 turn negative

Answer 4

because our general equation is this:
\[(x-h)^2 + (y -k)^2 = r^2 , centre (h, k)\]

Answer 5

b- 4,-3

Answer 6

So when we compare it to:
\[(x+2)^2 + (y - 5)^2 = 9, centre (-2, 5) \]

Answer 7

c- (5,6)

Answer 8

d- -1,7

Answer 9

is that right?

Answer 10

OKaaayyy let's seee :)
A. centre (-2, 5)
B. centre (4, -3)
C. centre (5, 6)
D. centre (-1, 7)

Answer 11

excellent work :)!

Answer 12

thank you :)

Answer 13

d and a are the answers

Answer 14

they lie in the second quadrant

Answer 15

no problem :)
That's correct :D Wonderful!

Answer 16

thank you for your help!

Answer 17

you're welcome ^_^!

Answer 18

also,

Answer 19

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

Answer 20

when it asks for whats its equation, does it mean to fully work out the equation or just set it up?

Answer 21

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 :)

Answer 22

it would be
(x+2)^2(x-1)^2=3^2

Answer 23

choice a

Answer 24

yess :D That's correct

Answer 25

thank you :)

Answer 26

you're welcome :D!