## 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 5 months ago 5 months ago

1. MrNood Group Title

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 Group Title

i know..

3. surjithayer Group Title

complete the squares and equate the centre.

4. mathmale Group Title

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 Group Title

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 Group Title

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 Group Title

i think its a..?

9. stupidinmath Group Title

-8

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

11. surjithayer Group Title

$\left( x^2+ax+\left( \frac{ a }{2 } \right)^2 \right)+\left( y^2+by+\left( \frac{ b }{ 2 } \right)^2 \right)=-2+\left( \frac{ a }{ 2 } \right)^2+\left( \frac{ b }{2 } \right)^2$ $\left( x-\frac{ a }{2 } \right)^2+\left( y-\frac{ b }{ 2 } \right)^2=-2+\frac{ a^2 }{4 }+\frac{ b^2 }{4 }$ 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 Group Title

-16 - 8.

13. stupidinmath Group Title

oh... positive 8?

14. mathmale Group Title

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 Group Title

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 Group Title

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

17. mathmale Group Title

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 Group Title

Welcome to Math Whiz