Complete the square and identify the conic. X^2+y^2+4x-6y+1=0
To complete the square for this equation, we will group the x and y terms together:
x^2 + 4x + y^2 - 6y + 1 = 0
Next, we need to add and subtract the necessary terms to complete the squares for x and y:
(x^2 + 4x + 4) + (y^2 - 6y + 9) - 4 - 9 + 1 = 0
(x + 2)^2 + (y - 3)^2 - 12 + 1 = 0
(x + 2)^2 + (y - 3)^2 - 11 = 0
Now, we can identify the conic from the completed square equation. Since both the x and y terms are squared, and have the same sign with equal coefficients, and the constants are on the right side of the equation, we have a standard form equation of a circle.
Therefore, the conic is a circle with center at (-2, 3) and radius sqrt(11).