To complete the square, let's manipulate the equation by grouping the x terms and y terms separately:
X^2 + 4x + y^2 - 6y + 1 = 0
(X^2 + 4x) + (y^2 - 6y) + 1 = 0
(X^2 + 4x + 4) - 4 + (y^2 - 6y + 9) - 9 + 1 = 0
(X + 2)^2 - 3 + (y - 3)^2 - 9 + 1 = 0
(X + 2)^2 + (y - 3)^2 - 11 = 0
(X + 2)^2 + (y - 3)^2 = 11
The conic formed by the equation is B. (X+2)^2 + (y-3)^2 = 11, which is a circle with center (-2, 3) and radius sqrt(11).
Use complete the square and identify the conic.
X^2+y^2+4x-6y+1=0
A. (X-2)^2+(y+3)^2=2 square root 3
B. (X+2)^2+(y-3)^2=12
1 answer
1 answer