anonymous
  • anonymous
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
Mathematics
chestercat
  • chestercat
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

MrNood
  • MrNood
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'
anonymous
  • anonymous
i know..
anonymous
  • anonymous
complete the squares and equate the centre.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

mathmale
  • mathmale
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.
mathmale
  • mathmale
Hint: the equation of a circle with radius r and centered at (h,k) is\[(x-h)^{2}+(y-k)^{2}=r ^{2}\]. Hope this puts you on track towards finding a solution to this problem.
mathmale
  • mathmale
Please demo how you'd complete the square: x^2 + 6x
RadEn
  • RadEn
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
anonymous
  • anonymous
i think its a..?
anonymous
  • anonymous
-8
RadEn
  • RadEn
have you solve them : -a/2 = 4 a = ... -b/2 = -8 b = ...
anonymous
  • anonymous
\[\left( x^2+ax+\left( \frac{ a }{2 } \right)^2 \right)+\left( y^2+by+\left( \frac{ b }{ 2 } \right)^2 \right)=-2+\left( \frac{ a }{ 2 } \right)^2+\left( \frac{ b }{2 } \right)^2\] \[\left( x-\frac{ a }{2 } \right)^2+\left( y-\frac{ b }{ 2 } \right)^2=-2+\frac{ a^2 }{4 }+\frac{ b^2 }{4 }\] centre is \[\left( \frac{ -a }{ 2 } ,\frac{ -b }{2 }\right)\] \[\frac{ -a }{ 2 }=4,a=-8,\frac{ -b }{2 }=-8,b=-8*-2=16\]
anonymous
  • anonymous
-16 - 8.
anonymous
  • anonymous
oh... positive 8?
mathmale
  • mathmale
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.
anonymous
  • anonymous
we can also find by comparing with \[x^2+y^2+2gx+2fy+c=0,where~centre~is~\left( -g,-f \right) ~and radius=\sqrt{g^2+f^2-c}\]
anonymous
  • anonymous
radius=\[\sqrt{g^2+f^2-c}\]
mathmale
  • mathmale
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?
MrNood
  • MrNood
Welcome to Math Whiz

Looking for something else?

Not the answer you are looking for? Search for more explanations.