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 11 months ago
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
 11 months ago
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

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MrNood
 11 months ago
Best ResponseYou've already chosen the best response.0It 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'

surjithayer
 11 months ago
Best ResponseYou've already chosen the best response.1complete the squares and equate the centre.

mathmale
 11 months ago
Best ResponseYou've already chosen the best response.0I'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.

mathmale
 11 months ago
Best ResponseYou've already chosen the best response.0Hint: 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.

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

RadEn
 11 months ago
Best ResponseYou've already chosen the best response.0just 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

stupidinmath
 11 months ago
Best ResponseYou've already chosen the best response.0i think its a..?

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

surjithayer
 11 months ago
Best 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 \[\left( \frac{ a }{ 2 } ,\frac{ b }{2 }\right)\] \[\frac{ a }{ 2 }=4,a=8,\frac{ b }{2 }=8,b=8*2=16\]

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

mathmale
 11 months ago
Best ResponseYou've already chosen the best response.0Dear 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.

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

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

mathmale
 11 months ago
Best ResponseYou've already chosen the best response.0I 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?
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