What is the standard form of the equation of the circle with center (-2, -3) and a radius of SQR OF 2/4?
@asnaseer
we just did the standard equation of a circle - remember a circle centered at (a,b) with radius r has equation:\[(x-a)^2+(y-b)^2=r^2\]
\[(X+3)^2+(Y+2)^2=1/8\]
VERY close - remember they said the radius is \(\sqrt{\frac{2}{4}}=\sqrt{\frac{1}{2}}\)
:(, WHAT DID I DO WRONG? I DONT UNDERSTAND
I didn't mean to do all caps
lol - don't worry :)
you are given\[r=\sqrt{\frac{2}{4}}=\sqrt{\frac{1}{2}}\]therefore \(r^2=?\)
hey @flowey15
you also have the X/Y swapped
ohh,
omg im sorry, we have been righting the radius|dw:1458059848216:dw| wrong
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
\[(x+2)^2+(y+3)^2=1/8\]
bingo! well done! :)
YAYY, THANKS SOOOO MUCH, YOUR THE BEST!!!!
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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