Question

Identify the center and radius of the circles represented by the following questions

(a) 3x² - √3x+3y²-y+ 1/12=0
(b) 12x² + 12y²+42y=0

Answer 1

first one:
divide each term by 3

x^2 -√3/3 + y^2 + y/3 = -1/36
add the terms to complete the square

x^2 -√3/3 + <b>1/12</b> + y^2 + y/3 + <b>1/36</b>= -1/36 + <b>1/12</b> + <b>1/36 </b>

(x - √3/6)^2 + (y - 1/6)^2 = 1/12

centre is (√3/6,1/6)
radius is 1/√12 = 1/2√3 = √3/6

You do the second one, it is easier.

Answer 2

Would you start by dividing everything by 6? then after that I am confused

Answer 3

no, by 12, you want to start the equation with 1x^2

Answer 4

Ok I started by diving by 12 to get

x² + y² + 3/2y = 0

then

x² + y² +3/2 + ____ = 0 + ____

then

(3/2)² = 3/4
(3/4)² = 9/16

and got the equation

x² + y² + (3/2)y + 9/16 = 0 + 9/16

here is where I am confused

x² + [y+(3/4)]² = (3/4)²

Answer 5

no

after you divide by 12 you should have had

x^2 + y^2 + (7/2)y = 0

to figure out what you have to add to both sides, take half of the middle term, then square that

so 1/2 of 7/2 = 7/4
and 7/4 squared is 49/16

x^2 + y^2 + (7/2)y + 49/16 = 0 + 49/16

x^2 + (y + 7/4)^2 = 49/16

centre: (0,-7/4)
radius: 7/4

Answer 6

the measures of two sides of a triangle are given the perimeter of the triangle is 13x2-14x+12 find the measure of the third side