Question

determine the centre and radius of the circle with equation 3x^2+3y^2-5x+6y+3=0

[using completion of squares]

please show all working...thank you :)

Answer 1

this is the same crap im doing lol good luck i have got no help

Answer 2

lol okay

Answer 3

This is the first part

Answer 4

(3x^2 -5x) + (3y^2 +6y) +3 = 0

3(x^2 -5/3 x) + 3(y^2 +2y) +3 = 0

3(x^2 -5/3x +(5/6)^2) + 3(y^2 +2y +1) +3-3((5)/6)^2-3(1) = 0

3(x -5/6)^2 + 3(y+1)^2 -3(25/36) = 0

3(x -5/6)^2 + 3(y+1)^2 -25/12 = 0 perhaps?

Answer 5

That is the second part

Answer 6

3(x -5/6)^2 + 3(y+1)^2 -25/12 = 0 ; \3

(x -5/6)^2 + (y+1)^2 -25/36 = 0

(x -5/6)^2 + (y+1)^2 = 25/36 maybe :)

Answer 7

center: (5/6 , -1)

radius = 5/6

Answer 8

Correct.

Answer 9

thanks emunrradtvamg and amistre but i dont see how you guys arrived at your final answer because of the genral from of the equation i saw how we got the radius because its r but k is the center?

Answer 10

Look at the second picture I send you. Im gonna try to explain

Answer 11

i look at it i saw the general form of the equation

(x-h)^2+(y-k)^2=r^2

Answer 12

Exactly

As you can see, inside of the first term you have (x-5/6)^2 that is equivalent to (x-h)^2. So you can realize from that that h = 5/6.

(y+1) is equivalent to (y-k)^2. So k = -1 (look carefully at the formula)

the center is (h,k) = (5/6,-1)

Answer 13

oh okay i get it now thank you so much emunrradtvamg and amistre!!!