Complete the square for this conic section in standard form:
x^2+y^2-6x+4y-12=0
x^2+y^2-6x+4y-12=0
x^2 - 6x + 9 + y^2 + 4y 4 = 12 + 9 + 4
(x-3)^2 + (y+2)^2 = 25
To complete the square for a conic section given in standard form, follow these steps:
Step 1: Group the x terms and the y terms separately.
x^2 - 6x + y^2 + 4y = 12
Step 2: For the x terms, take half of the coefficient of x (-6) and square it: (-6/2)^2 = 9.
To maintain the balance, add this value inside the parentheses, and subtract it outside the parentheses. Similarly, for the y terms, take half of the coefficient of y (4), square it: (4/2)^2 = 4. Add this value inside the parentheses, and subtract it outside the parentheses.
x^2 - 6x + 9 + y^2 + 4y + 4 = 12 + 9 + 4
Step 3: Simplify the equation by combining the constants.
x^2 - 6x + y^2 + 4y + 13 = 25
Step 4: Rearrange the equation by separately arranging the squared terms (x^2 and y^2), the x terms (-6x) and the y terms (4y), and the constants (13 and 25).
(x^2 - 6x) + (y^2 + 4y) = 12
Step 5: Rewrite the squared terms, x^2 - 6x, and y^2 + 4y, as perfect squares.
(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4
Step 6: Simplify the squared terms.
(x - 3)^2 + (y + 2)^2 = 25
The conic section x^2 + y^2 - 6x + 4y - 12 = 0 can be written in standard form as (x - 3)^2 + (y + 2)^2 = 25. This represents a circle with radius 5 units and center at (3, -2).