Identify the conic section whose equation is given:
x2 + 10x + y2 + 12y = 60
x^2 + 10 x + (10/2)^2 + y^2 +12 y + (12/2)^2 = 60 +(10/2)^2 + (12/2)^2
x^2 + 10 x + 25 + y^2 + 12 y + 36 = 60 + 25 + 36
(x+5)^2 + (y+6)^2 = 121
That is a circle with radius 11 centered at (-5,-6).
To identify the conic section, we can analyze the coefficients of the variables in the equation:
x^2 + 10x + y^2 + 12y = 60
1. Group the x-terms and y-terms together:
(x^2 + 10x) + (y^2 + 12y) = 60
2. Complete the square for both x and y separately:
- For the x-terms:
Take half of the coefficient of x (which is 10), square it, and add it to both sides of the equation:
(x^2 + 10x + 25) + (y^2 + 12y) = 60 + 25
(x + 5)^2 + (y^2 + 12y) = 85
- For the y-terms:
Take half of the coefficient of y (which is 12), square it, and add it to both sides of the equation:
(x + 5)^2 + (y^2 + 12y + 36) = 85 + 36
(x + 5)^2 + (y + 6)^2 = 121
The equation can now be written in the standard form:
(x + 5)^2 + (y + 6)^2 = 11^2
Comparing this equation to the standard form of different conic sections, we see that the equation resembles the standard form of a circle:
(x - h)^2 + (y - k)^2 = r^2
Therefore, the given equation represents a circle with a center at (-5, -6) and a radius of 11.