Question

What is the standard form of 3x^2+3y^2+12x+18y-15=0

Answer 1

Answer:

(x + 2)^2 + (y + 3)^2 = 18

Step-by-step explanation:

To write this equation in the standard form i.e. (x - h)^2 + (y - k)^2 = R^2, we need to complete the squares on the left side.

For completing the squares, put variables on one side and constants on the other side:

3x^2 + 3y^2 + 12x + 18y - 15 = 0

3x^2 + 3y^2 + 12x + 18y  = 15

Rearrange the like terms together:

3x^2 + 12x + 3y^2 + 18y  = 15

Since the coefficients of x^2 and y^2 terms are the same, divide both the sides by this coefficient i.e. 3 to get:

x^2 + 4x + y^2 + 6y  = 5

Now, square the 1/2 of coefficients for x and y and add them to both sides of the equation:

square of 1/2 of coefficient for x = (4/2)^2 = 4; and

square the 1/2 of coefficient for y = (6/2)^2 = 9

x^2 + 4x + 4 + y^2 + 6y + 9 = 5 + 4 + 9

x^2 + 4x + 4 + y^2 + 6y + 9 = 18

Rewriting this as perfect squares:

(x + 2)^2 + (y + 3)^2 = 18