# How do you identity if the equation 2x^2+12x+18-y^2=3(2-y^2)+4y is a parabola, circle, ellipse, or hyperbola and how do you graph it?

Nov 15, 2016

Set your compass to a radius of 2, put the center at the point $\left(- 3 , 1\right)$, and draw a circle.

#### Explanation:

Here is a helpful reference Conic Section - General form

Combine all of the terms to be in the general form:

$A {x}^{2} + B x y + C {y}^{2} + D x + E y + F = 0$

$2 {x}^{2} + 0 x y + 2 {y}^{2} + 12 x - 4 y + 12 = 0 \text{ }$

Because $A = C \mathmr{and} B = 0$, this is an equation of a circle. The general equation for a circle is:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

where $\left(x , y\right)$ is any point on the circle, $\left(h , k\right)$ is the center, and r is the radius.

Divide both sides of equation  by 2:

${x}^{2} + {y}^{2} + 6 x - 2 y + 6 = 0 \text{ }$

Add ${h}^{2} + {k}^{2} - 6$ to both sides of the equation:

${x}^{2} + 6 x + {h}^{2} + {y}^{2} - 2 y + {k}^{2} = {h}^{2} + {k}^{2} - 6 \text{ }$

Set the term middle in the right side of the pattern, ${\left(x - h\right)}^{2} = {x}^{2} - 2 x h + {h}^{2}$, equal to the corresponding term in equation :

$- 2 h x = 6 x$

Solve for h:

$h = - 3$

Substitute the left side on the pattern into equation 3:

${\left(x - h\right)}^{2} + {y}^{2} - 2 y + {k}^{2} = {h}^{2} + {k}^{2} - 6 \text{ }$

Substitute -3 for h in equation :

${\left(x - - 3\right)}^{2} + {y}^{2} - 2 y + {k}^{2} = {\left(- 3\right)}^{2} + {k}^{2} - 6 \text{ }$

Set the term middle in the right side of the pattern, ${\left(y - k\right)}^{2} = {y}^{2} - 2 y k + {k}^{2}$, equal to the corresponding term in equation :

$- 2 y k = - 2 y$

Solve for k:

$k = 1$

Substitute the left side of the pattern into equation :

${\left(x - - 3\right)}^{2} + {\left(y - k\right)}^{2} = {\left(- 3\right)}^{2} + {k}^{2} - 6 \text{ }$

Substitute 1 for k into equation :

${\left(x - - 3\right)}^{2} + {\left(y - 1\right)}^{2} = {\left(- 3\right)}^{2} + {1}^{2} - 6 \text{ }$

Simplify the constant terms:

${\left(x - - 3\right)}^{2} + {\left(y - 1\right)}^{2} = 4 \text{ }$

Write the constant term as a square:

${\left(x - - 3\right)}^{2} + {\left(y - 1\right)}^{2} = {2}^{2} \text{ }$

Equation  is the standard form of a circle with a center at $\left(- 3 , 1\right)$ and a radius of 2.