What is the standard form of the equation of a circle that has its center at (-2, -3) and passes through the point (-2, 0)?

(x + 2)^2 + (y + 3)^2 = 9

(x − 2)^2 + (y − 3)^2 = 16

(x − 2)^2 + (y − 3)^2 = 4

(x + 2)^2 + (y + 3)^2 = 16

(x − 2)2 + (y + 3)2 = 9

Question

What is the standard form of the equation of a circle that has its center at (-2, -3) and passes through the point (-2, 0)?

    (x + 2)^2 + (y + 3)^2 = 9

    (x − 2)^2 + (y − 3)^2 = 16

    (x − 2)^2 + (y − 3)^2 = 4

    (x + 2)^2 + (y + 3)^2 = 16

    (x − 2)2 + (y + 3)2 = 9

undeadknight26 · Answer

Hewo der rainbow!  Which do you think it is?

anonymous · Answer

I honestly have no clue what so ever, but if I had to guess I would choose the last one.

misty1212 · Answer

JI!!

undeadknight26 · Answer

Close...But not right...

misty1212 · Answer

no not the last one don't pick that one

undeadknight26 · Answer

Last one is horrible choice ^

anonymous · Answer

Lol, i told you, I have no clue

undeadknight26 · Answer

Probably @misty1212 can tell you how to do it...I just graph it :p

misty1212 · Answer

$$(x-h)^2+(y-k)^2=r^2$$ and in your case $(h,k)$ is $(-2,-3)$ so 
$$(x+2)^2+(y+3)^2=r^2$$

anonymous · Answer

Okay, so how would I find the r^2?

misty1212 · Answer

$r$ is the distance between $(-2,0)$ and $(-2,3)$ which is evidently $3$ so $$(x+2)^2+(y+3)^2=3^2$$

anonymous · Answer

Ohhh I get it, so the answer is $$(x+2)^2 + (y+3)^2 = 9 $$

undeadknight26 · Answer

Yup.

anonymous · Answer

Ohh, okay thanks to both. That really helped me :)