circle theorem? Give the center and radius of (x-3)^2 + (y+2)^2 =8

[ect]

Give the center and radius of (x-3)² [FONT=Times New Roman, Times, serif]+ (y+2)[/FONT]² [FONT=Times New Roman, Times, serif]=8. [/FONT]
[FONT=Times New Roman, Times, serif]I dont quit understand what to do because the formula is (x-h)[/FONT]² [FONT=Times New Roman, Times, serif]+ (y-k)[/FONT]²[FONT=Times New Roman, Times, serif] = r[/FONT]²[FONT=Times New Roman, Times, serif] , I know that (3,-2) is the center, but 8 doesn't have a perfect square and so I tried doing the square root of 4 times 2 which is 2times the square root of 2 but thats wrong aswell and I have to graph it so I don't think its a number with decimals.[/FONT]

 

[ksdhart2]

Are you perhaps entering these answers on an online platform? I only ask because it's well known that such systems are often very difficult to work with and mark correct answers are wrong because they don't match the exact, extremely particular, form it's looking for. The radius of the circle in question is the square root of 8, as you identified. You might try entering sqrt(8) instead of 2sqrt(2), or possibly even try 2*sqrt(2). As stupid as it sounds, I have seen online answer checking systems reject solutions because of a missing (implicit) times sign. As for graphing the circle, the radius is an irrational number, so any graph you draw will be an approximation, so I wouldn't fret too much about it being off. Just do the best you can.

 

[pka]

Give the center and radius of (x-3)² + (y+2)² =8.
I dont quit understand what to do because the formula is (x-h)² + (y-k)² = r² , I know that (3,-2) is the center, but 8 doesn't have a perfect square and so I tried doing the square root of 4 times 2 which is 2times the square root of 2 but thats wrong aswell and I have to graph it so I don't think its a number with decimals.

Have you tried just \(\displaystyle \sqrt{8}~?\)

 

[Steven G]

As for graphing the circle, the radius is an irrational number, so any graph you draw will be an approximation, so I wouldn't fret too much about it being off. Just do the best you can.

I disagree with that. You get to use your own scale which can be in units of sqrt(2)'s! In that case you will have difficulty finding the center (3, -2) because they are rational numbers!!

 

[ect]

wow. I thought I had tried the square root of 8. I guess I didn't think of it because the system is a little weird sometimes it just marks it wrong because its not in the right form or int a decimal answer or it needs to be rounded or it needs parenthesis or something. thank you!

 

[pka]

wow. I thought I had tried the square root of 8. I guess I didn't think of it because the system is a little weird sometimes it just marks it wrong because its not in the right form or int a decimal answer or it needs to be rounded or it needs parenthesis or something. thank you!

After working with testing companies, I understand that their contract programmers are lazy. The easy way is to allow only one form of the answer. In this case, although \(\displaystyle 2\sqrt 2=\sqrt 8\) only the second from is correct because it happens to be the most obvious.