Asked by Don
the center of a circle
x^2 + y^2 + 8y =1 is
a. (0,-2)
b. (0,4)
c. (0,1)
d. (0,-4)
I believe the answer is (0,4)
Am I correct ?
x^2 + y^2 + 8y =1 is
a. (0,-2)
b. (0,4)
c. (0,1)
d. (0,-4)
I believe the answer is (0,4)
Am I correct ?
Answers
Answered by
Reiny
Incorrect!
you have to complete the square to get it into standard form
x^2 + y^2 + 8y <b>+ 16 </b>= 1 <b>+ 16 </b>
x^2 + (y+4)^2 = 17
Now what do you know from your previous posts on this?
you have to complete the square to get it into standard form
x^2 + y^2 + 8y <b>+ 16 </b>= 1 <b>+ 16 </b>
x^2 + (y+4)^2 = 17
Now what do you know from your previous posts on this?
Answered by
Don
I'm not sure what your looking for Reiny. It looks like the center of the circle is still (0,4) what am I doing wrong?
Answered by
Don
(x+0)^2 + (y+4)^2 = 17
the center would be (0,4)
I can't see it any other way ...arghhhh
the center would be (0,4)
I can't see it any other way ...arghhhh
Answered by
Reiny
I pointed out the pattern for you in
http://www.jiskha.com/display.cgi?id=1325732520
and you seem to be able to understand in your posts that follow it,
Now suddenly you have it backwards again.
remember that the sign is opposite what you see in the brackets
so for (x-5)^2 + (y+7)^2 = 100
the centre would be (+5, -7)
so for the above question is would obviously be
(0, -4)
http://www.jiskha.com/display.cgi?id=1325732520
and you seem to be able to understand in your posts that follow it,
Now suddenly you have it backwards again.
remember that the sign is opposite what you see in the brackets
so for (x-5)^2 + (y+7)^2 = 100
the centre would be (+5, -7)
so for the above question is would obviously be
(0, -4)
Answered by
Don
ARGHHHHH now that really was a brain fart on my part ...
Thanks a lot Reiny for the help
Don
Thanks a lot Reiny for the help
Don
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