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stupidinmath
Group Title
If the center of the circle x^2 + y^2 + ax + by + 2 = 0 is point (4,8), what is a + b?
a. 8
b. 4
c. 4
d. 8
e. 24
 8 months ago
 8 months ago
stupidinmath Group Title
If the center of the circle x^2 + y^2 + ax + by + 2 = 0 is point (4,8), what is a + b? a. 8 b. 4 c. 4 d. 8 e. 24
 8 months ago
 8 months ago

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MrNood Group TitleBest ResponseYou've already chosen the best response.0
It is rare in life to find someone who would call themselves 'stupid in English' or 'stupid in sport' Maths is only another skill to be learned and enjoyed like any other learning. It is not a badge of pride to 'not be able to do maths'
 8 months ago

stupidinmath Group TitleBest ResponseYou've already chosen the best response.0
i know..
 8 months ago

surjithayer Group TitleBest ResponseYou've already chosen the best response.1
complete the squares and equate the centre.
 8 months ago

mathmale Group TitleBest ResponseYou've already chosen the best response.0
I'd go so far as to try opening a new account with OpenStudy, so that you don't go on labeling yourself with such a derogatory term. It's not funny and does you no good.
 8 months ago

mathmale Group TitleBest ResponseYou've already chosen the best response.0
Hint: the equation of a circle with radius r and centered at (h,k) is\[(xh)^{2}+(yk)^{2}=r ^{2}\]. Hope this puts you on track towards finding a solution to this problem.
 8 months ago

mathmale Group TitleBest ResponseYou've already chosen the best response.0
Please demo how you'd complete the square: x^2 + 6x
 8 months ago

RadEn Group TitleBest ResponseYou've already chosen the best response.0
just an alternative : if given the equation of circle x^2 + y^2 + ax + by + c = 0, then the centre is (a/2 , b/2). knowed the centre is (4,  8) so, a/2 = 4 solve for a also b/2 = 8 solve for b after that, you can determine the value of a + b
 8 months ago

stupidinmath Group TitleBest ResponseYou've already chosen the best response.0
i think its a..?
 8 months ago

RadEn Group TitleBest ResponseYou've already chosen the best response.0
have you solve them : a/2 = 4 a = ... b/2 = 8 b = ...
 8 months ago

surjithayer Group TitleBest ResponseYou've already chosen the best response.1
\[\left( x^2+ax+\left( \frac{ a }{2 } \right)^2 \right)+\left( y^2+by+\left( \frac{ b }{ 2 } \right)^2 \right)=2+\left( \frac{ a }{ 2 } \right)^2+\left( \frac{ b }{2 } \right)^2\] \[\left( x\frac{ a }{2 } \right)^2+\left( y\frac{ b }{ 2 } \right)^2=2+\frac{ a^2 }{4 }+\frac{ b^2 }{4 }\] centre is \[\left( \frac{ a }{ 2 } ,\frac{ b }{2 }\right)\] \[\frac{ a }{ 2 }=4,a=8,\frac{ b }{2 }=8,b=8*2=16\]
 8 months ago

stupidinmath Group TitleBest ResponseYou've already chosen the best response.0
16  8.
 8 months ago

stupidinmath Group TitleBest ResponseYou've already chosen the best response.0
oh... positive 8?
 8 months ago

mathmale Group TitleBest ResponseYou've already chosen the best response.0
Dear Math Whiz: Why not try doing some of these calculations yourself and then sharing the results? Those trying to help you could then give you more specific and useful feedback.
 8 months ago

surjithayer Group TitleBest ResponseYou've already chosen the best response.1
we can also find by comparing with \[x^2+y^2+2gx+2fy+c=0,where~centre~is~\left( g,f \right) ~and radius=\sqrt{g^2+f^2c}\]
 8 months ago

surjithayer Group TitleBest ResponseYou've already chosen the best response.1
radius=\[\sqrt{g^2+f^2c}\]
 8 months ago

mathmale Group TitleBest ResponseYou've already chosen the best response.0
I still think completing the square is a more readily understandable approach. But first, Math Whiz (I refuse to call you stupidinmath), are you familiar with completing the square?
 8 months ago

MrNood Group TitleBest ResponseYou've already chosen the best response.0
Welcome to Math Whiz
 8 months ago
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