SOLUTION: Graph the equation 2x^2 + 2y^2 -16x + 4y

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Question 386132: Graph the equation
2x^2 + 2y^2 -16x + 4y - 38 = 0
Found 3 solutions by ewatrrr, solver91311, jsmallt9:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!

Hi,
Standard Form of an Equation of a Circle is
where Pt(h,k) is the center.
Graphing
2x^2 + 2y^2 -16x + 4y - 38 = 0 factoring out the 2
x^2 + y^2 -8x + 2y - 19 = 0
x^2 -8x + y^2 + 2y - 19 = 0 Compleitn both squares
(x-4)^2 -16 + (y +1)^2 - 1 -19 = 0
(x-4)^2 + (y +1)^2 = 36
Center is (4,-1) radius = 6

Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!

First, what is it? 2nd degree equation in two variables. Has to be a conic section.

Has both and terms -- NOT a parabola.

and terms have the same sign -- NOT a hyperbola.

The coefficients on the and terms are equal -- NOT an ellipse.

Has to be a circle.

Find the center and radius. Complete the square on each of the two variables.

Move the constant term into the RHS:

Group same type variables:

Divide by the lead coefficient:

Divide the coefficient on th 1st degree x term by 2, square the result, add that result to both sides.

Repeat for the coefficient on the 1st degree y term:

Factor the two perfect square trinomials in the LHS and simplify the RHS:

Hence you have a circle centered at (4,-1) with radius 6.

John

My calculator said it, I believe it, that settles it

Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website!

Since everything is divisible by 2, I am going to start by dividing both sides by 2. This will create smaller numbers and coefficients of 1 in front of the squared terms. This will make the rest of the problem easier:

With no xy term and with equal coefficients in front of the squared terms, this equation is the equation of a circle. So we want it in the general form for the equation of a circle:

So we need to complete the squares for both the x terms and the y terms. When completing squares I like to start by "moving" the constant term to the other side of the equation. So I'll add 19 to each side (and rearrange the terms so the x terms and y terms are together):

The next step is to figure out what constant terms are needed to make
a perfect square trinomial, and
a perfect square trinomial.

To find these constant terms we take half of the coeeficient of x (or y) and square it. Half of -8 is -4. -4 squared is 16. So we need to add 16 to each side to complete the square for the x terms. For the y terms, half of 2 is 1 and 1 squared is 1 so we need to add 1 to each side to complete the square for the y terms:

On the left side we can rewrite the equation as two perfect squares and on the right side we just add up the numebers:

The only things left to do are to write the perfect square for the y's as a subtraction and to write the right side as a perfect square:

Now the equation is in the desired form for a circle. We can read the h and the k which are coordinates of the center of the circle and the r which is the radius:
h = 4
k = -1
Center: (4, -1)
r = radius = 6
With the center and the radius you should be able to sketch a graph of this circle.

SOLUTION: Graph the equation 2x^2 + 2y^2 -16x + 4y - 38 = 0