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
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 + 1/12 + y^2 + y/3 + 1/36= -1/36 + 1/12 + 1/36
(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.
Would you start by dividing everything by 6? then after that I am confused
no, by 12, you want to start the equation with 1x^2
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
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
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
To identify the center and radius of a circle represented by an equation, we need to rewrite the equations in standard form, which is given by (x - h)² + (y - k)² = r², where (h, k) represents the center of the circle and r represents the radius.
Let's solve each equation step by step:
(a) 3x² - √3x + 3y² - y + 1/12 = 0
To rewrite it in standard form, we need to complete the squares for both x and y terms:
Rearrange the equation:
3x² - √3x + 3y² - y = -1/12
For x terms:
First, divide the coefficient of x by 2 and square it: (√3/2)² = 3/4
Add the result to both sides of the equation:
3x² - √3x + (3/4) = -1/12 + (3/4) = 8/12 = 2/3
For y terms:
First, divide the coefficient of y by 2 and square it: (-1/2)² = 1/4
Add the result to both sides of the equation:
3x² - √3x + (3/4) + 3y² - y + (1/4) = 2/3 + 1/4 = 11/12
Simplify the equation:
3x² - √3x + 3y² - y + 1 = 11/12
Now we can rewrite the equation in standard form:
3x² - √3x + 3y² - y = 11/12 - 1 = 12/12
3x² - √3x + 3y² - y = 1
Comparing it to the standard form equation (x - h)² + (y - k)² = r², we can deduce that:
Center of the circle = (h, k) = (1/2√3, 1/6)
Radius of the circle = r = √(1/2² + 1/6²) = √(1/4 + 1/36) = √(9/36) = 3/6 = 1/2
Therefore, the center of the circle is (1/2√3, 1/6) and the radius is 1/2.
(b) 12x² + 12y² + 42y = 0
To rewrite it in standard form, we need to complete the squares for both x and y terms:
First, let's divide the equation by 12 to simplify it: x² + y² + 7y = 0
For x terms:
Add (1/2)² = 1/4 to both sides of the equation:
x² + (1/4) + y² + 7y = 1/4
For y terms:
Add (7/2)² = 49/4 to both sides of the equation:
x² + (1/4) + y² + 7y + 49/4 = 1/4 + 49/4 = 50/4 = 25/2
Simplify the equation:
x² + y² + (7/2)y + 1/4 = 25/2
Now we can rewrite the equation in standard form:
x² + y² + (7/2)y = 25/2 - 1/4 = 48/4 - 1/4 = 47/4
Comparing it to the standard form equation (x - h)² + (y - k)² = r², we can deduce that:
Center of the circle = (h, k) = (0, -7/4)
Radius of the circle = r = √(47/4)
Therefore, the center of the circle is (0, -7/4) and the radius is √(47/4).