Question

52. What is the standard equation of the circle in the graph?
(x + 3)2 + (y – 2)2 = 9
(x – 3)2 + (y + 2)2 = 9
(x – 3)2 + (y + 2)2 = 3
(x + 3)2 + (y – 2)2 = 3

Answer 1

wheres the circle

Answer 2

tell me the center of the circle.

Answer 3

\(\Large\color{red}{ \bf (x-h)^2+(y-k)^2=r^2 }\)

Answer 4

I dont know the center. I dont understand this one at all

Answer 5

I think the center is -2,2

Answer 6

what's the x-coordinate of the circle you think?

Answer 7

The center is -3,2

Answer 8

yeap.... that's about right so now we know the center of the circle (-3, 2) or (h,k)

so h=-3 and k=2

any ideas on what the radius might be?

Answer 9

so the radius is 2?

Answer 10

is it? look at the picture... how "long" is the radius?

Answer 11

ohhh 3?

Answer 12

well, then now you know what "h" and "k" and "r" are, so just plug them in the circle equation :)

Answer 13

huhh? Im so lost. Ive been doing this test since 11 this morning

Answer 14

I think its c?

Answer 15

\(\bf (x-h)^2+(y-k)^2=r^2 \qquad
\begin{cases}
h=-3\\
\bf k=2\\
\bf r=3
\end{cases}
\\ \quad \\
\implies (x-(-3))^2+(y-(-2))^2=(3)^2\)

Answer 16

hmmm actaully shoot, I got an extra minus there =| \(\bf (x-h)^2+(y-k)^2=r^2 \qquad
\begin{cases}
h=-3\\
\bf k=2\\
\bf r=3
\end{cases}
\\ \quad \\
\implies (x-(-3))^2+(y-(2))^2=(3)^2\)

Answer 17

so... what would you get?

Answer 18

okay so d?

Answer 19

is it?

Answer 20

is that what you got?

Answer 21

Thats what i got, I just dont know if its right tho

Answer 22

\(\bf (x-(-3))^{\color{red}{ 2}}+(y-(-2))^{\color{red}{ 2}}=(3)^{\color{red}{ 2}}\)

Answer 23

man another typo.... shoot

Answer 24

\(\bf (x-(-3))^{\color{red}{ 2}}+(y-(2))^{\color{red}{ 2}}=(3)^{\color{red}{ 2}}\)

Answer 25

how did you get D anyway?

Answer 26

Nevermind I got a. I just figured it out i think