anonymous 5 years ago Can you help me to find the center and the radio of the following circle. (2x^2)+(2y^2)-8x+12y-28=0

1. anonymous

2. anonymous

is dis d ans tel

3. anonymous

dont know. Can you teach me how to? Dont care about the answer

4. anonymous

step by step

5. anonymous

k first divide eqn by 2 so that there is no coefficient of x^2 and y^2

6. anonymous

what if I have a 2 in x^2 and a 3 in y^2?

7. anonymous

i dont thnk thats possible bcoz coefficient of x^2 is always equal to coefficient of y^2 in a circle

8. anonymous

nice. So after dividing the equation by 2 I got: x^2+y^2-4x+y-14=0 what else should i do?

9. anonymous

then compar with eqn x^2 +y^2 +2gfx+2hy+c=0 we see -4x=2gfx n 6y=2hy g=-2 n h=3 centre(-g,-h) (2,-3) n radius r^2=g^2 +h^2-c r^2=4+9+14 r^2=27 r=3$\sqrt{3}$

10. anonymous

got it

11. anonymous

checking it out :) till now yes

12. anonymous

Ok in the third line you state that -4x=2gfx and in the next g=-2. how do you get rid of f?

13. anonymous

srry there is no f remove wherever i hav used

14. anonymous

ok

15. anonymous

hapy it helpd u

16. anonymous

According to what I understand, we have to leave x^2 and y^2 with coefficient 1. Is that correct?

17. anonymous

Then we have to match the equation with x^2 +y^2 +2gfx+2hy+c=0.

18. anonymous

find the values of g and h and we get the radius.

19. anonymous

sorry the center