What are the coordinates and radius of the circle whose equation is 2x raise to power 2 plus 2y raise to power 2 plus 4x minus 12y plus 12 equal to zero .
AAAaannndd the bot gets it wrong yet again!
2x raise to power 2 plus 2y raise to power 2 plus 4x minus 12y plus 12
2x^2+4x + 2y^2-12y = -12
(x+1)^2 + (y-3)^2 = 2^2
Why are you posting this question is the same silly typed way even
after oobleck told you to type it in the normal way??
Also, "What are the coordinates and radius of the circle.." should have said:
What are the coordinates of the centre and the radius of the circle ....
2x raise to power 2 plus 2y raise to power 2 plus 4x minus 12y plus 12 equal to zero
2x^2 + 2y^2 + 4x - 12y - 12 = 0
Add 12 to both sides, then divide each term by 2
x^2 + 2x + y^2 - 6y = 6
complete the square, by adding numbers as show in my next line
x^2 + 2x + 1 + y^2 - 6y + 9 = 6+1+9
(x+1)^2 + (y-3)^2 = 16
then centre is (-1,3) and the radius is 4
btw, as you can see the bot is WRONG again
To find the center and radius of a circle given its equation in standard form, you can follow these steps:
Step 1: Rewrite the equation in the form (x - h)^2 + (y - k)^2 = r^2.
Step 2: Identify the values of h, k, and r.
Let's apply these steps to the given equation: 2x^2 + 2y^2 + 4x - 12y + 12 = 0.
Step 1: Complete the square for the x terms:
2x^2 + 4x + k + 2y^2 - 12y + 12 = 0
2(x^2 + 2x) + 2y^2 - 12y + 12 = 0
2(x^2 + 2x + 1) + 2y^2 - 12y + 12 = 2(1)
2(x + 1)^2 + 2y^2 - 12y + 12 = 2
Step 1 (continued): Complete the square for the y terms:
2(x + 1)^2 + 2(y^2 - 6y) + 12 = 2
2(x + 1)^2 + 2(y^2 - 6y + 9) + 12 = 2 + 2(9)
2(x + 1)^2 + 2(y - 3)^2 + 12 = 20
Divide the equation by 2:
(x + 1)^2 + (y - 3)^2 + 6 = 10
Now, we have the equation in the desired form: (x - h)^2 + (y - k)^2 = r^2.
Step 2: Compare the equation to the standard form and identify the values of h, k, and r:
From the equation (x + 1)^2 + (y - 3)^2 + 6 = 10, we can identify the following:
h = -1
k = 3
r^2 = 10 - 6 = 4
r = 2 (taking the square root of both sides)
Therefore, the center of the circle is (-1, 3), and the radius is 2.