## stupidinmath one year 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

1. 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'

2. stupidinmath

i know..

3. surjithayer

complete the squares and equate the centre.

4. 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.

5. 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.

6. mathmale

Please demo how you'd complete the square: x^2 + 6x

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. stupidinmath

i think its a..?

9. stupidinmath

-8

have you solve them : -a/2 = 4 a = ... -b/2 = -8 b = ...

11. surjithayer

$\left( x^2+ax+\left( \frac{ a }{2 } \right)^2 \right)+\left( y^2+by+\left( \frac{ b }{ 2 } \right)^2 \right)=-2+\left( \frac{ a }{ 2 } \right)^2+\left( \frac{ b }{2 } \right)^2$ $\left( x-\frac{ a }{2 } \right)^2+\left( y-\frac{ b }{ 2 } \right)^2=-2+\frac{ a^2 }{4 }+\frac{ b^2 }{4 }$ centre is $\left( \frac{ -a }{ 2 } ,\frac{ -b }{2 }\right)$ $\frac{ -a }{ 2 }=4,a=-8,\frac{ -b }{2 }=-8,b=-8*-2=16$

12. stupidinmath

-16 - 8.

13. stupidinmath

oh... positive 8?

14. 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.

15. surjithayer

we can also find by comparing with $x^2+y^2+2gx+2fy+c=0,where~centre~is~\left( -g,-f \right) ~and radius=\sqrt{g^2+f^2-c}$

16. surjithayer

radius=$\sqrt{g^2+f^2-c}$

17. 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?

18. MrNood

Welcome to Math Whiz