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. 

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