Question

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

Answer 1

CENTRE IS (2,-3) radius squareroot(42)

Answer 2

is dis d ans tel

Answer 3

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

Answer 4

step by step

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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\[\sqrt{3}\]

Answer 10

got it

Answer 11

checking it out :) till now yes

Answer 12

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

Answer 13

srry there is no f remove wherever i hav used

Answer 14

ok

Answer 15

hapy it helpd u

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

sorry the center