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

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

Mathematics
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CENTRE IS (2,-3) radius squareroot(42)
is dis d ans tel
dont know. Can you teach me how to? Dont care about the answer

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step by step
k first divide eqn by 2 so that there is no coefficient of x^2 and y^2
what if I have a 2 in x^2 and a 3 in y^2?
i dont thnk thats possible bcoz coefficient of x^2 is always equal to coefficient of y^2 in a circle
nice. So after dividing the equation by 2 I got: x^2+y^2-4x+y-14=0 what else should i do?
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\[\sqrt{3}\]
got it
checking it out :) till now yes
Ok in the third line you state that -4x=2gfx and in the next g=-2. how do you get rid of f?
srry there is no f remove wherever i hav used
ok
hapy it helpd u
According to what I understand, we have to leave x^2 and y^2 with coefficient 1. Is that correct?
Then we have to match the equation with x^2 +y^2 +2gfx+2hy+c=0.
find the values of g and h and we get the radius.
sorry the center

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