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Mathematics 12 Online
OpenStudy (flowey15):

What is the standard form of the equation of the circle with center (-2, -3) and a radius of SQR OF 2/4?

OpenStudy (flowey15):

@asnaseer

OpenStudy (asnaseer):

we just did the standard equation of a circle - remember a circle centered at (a,b) with radius r has equation:\[(x-a)^2+(y-b)^2=r^2\]

OpenStudy (flowey15):

\[(X+3)^2+(Y+2)^2=1/8\]

OpenStudy (asnaseer):

VERY close - remember they said the radius is \(\sqrt{\frac{2}{4}}=\sqrt{\frac{1}{2}}\)

OpenStudy (flowey15):

:(, WHAT DID I DO WRONG? I DONT UNDERSTAND

OpenStudy (flowey15):

I didn't mean to do all caps

OpenStudy (asnaseer):

lol - don't worry :)

OpenStudy (asnaseer):

you are given\[r=\sqrt{\frac{2}{4}}=\sqrt{\frac{1}{2}}\]therefore \(r^2=?\)

OpenStudy (james1107):

hey @flowey15

OpenStudy (asnaseer):

you also have the X/Y swapped

OpenStudy (flowey15):

ohh,

OpenStudy (flowey15):

omg im sorry, we have been righting the radius|dw:1458059848216:dw| wrong

OpenStudy (asnaseer):

ohhh - ok, so radius \(r=\frac{\sqrt{2}}{4}\) which means \(r^2=\frac{2}{16}=\frac{1}{8}\) so you got that part correct :) you just need to fix the X/Y parts

OpenStudy (flowey15):

\[(x+2)^2+(y+3)^2=1/8\]

OpenStudy (asnaseer):

bingo! well done! :)

OpenStudy (flowey15):

YAYY, THANKS SOOOO MUCH, YOUR THE BEST!!!!

OpenStudy (asnaseer):

you are more than welcome my friend - maths is mostly about practice - the more you practice the better you'll get :)

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