SOLUTION: -3x^2 - 3y^2 + 12x + 96=0 a. find the center b. find the radius

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Question 609019: -3x^2 - 3y^2 + 12x + 96=0
a. find the center
b. find the radius

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 
Hi
-3x^2 - 3y^2 + 12x + 96=0
x^2 + y^2 - 4x - 32 =0 ||multiplying thru by -1/3
(x-2)^2 -4 +(y)^2 -32 =0 || completing the Square
%28x-2%29%5E2+%2B%28y%29%5E2+=+36+ |C(2,0 ) and r = 6
Note:
Standard Form of an Equation of a Circle is %28x-h%29%5E2+%2B+%28y-k%29%5E2+=+r%5E2
where Pt(h,k) is the center and r is the radius

See below descriptions of various conics
_______________________________________________________________________
Standard Form of an Equation of a Circle is %28x-h%29%5E2+%2B+%28y-k%29%5E2+=+r%5E2
where Pt(h,k) is the center and r is the radius
Standard Form of an Equation of an Ellipse is %28x-h%29%5E2%2Fa%5E2+%2B+%28y-k%29%5E2%2Fb%5E2+=+1+ where Pt(h,k) is the center. (a positioned to correspond with major axis)
a and b are the respective vertices distances from center and ±sqrt%28a%5E2-b%5E2%29are the foci distances from center: a > b
Standard Form of an Equation of an Hyperbola opening right and left is:
%28x-h%29%5E2%2Fa%5E2+-+%28y-k%29%5E2%2Fb%5E2+=+1 where Pt(h,k) is a center with vertices 'a' units right and left of center.
Standard Form of an Equation of an Hyperbola opening up and down is:
%28y-k%29%5E2%2Fb%5E2+-+%28x-h%29%5E2%2Fa%5E2+=+1 where Pt(h,k) is a center with vertices 'b' units up and down from center.
the vertex form of a parabola opening up or down, y=a%28x-h%29%5E2+%2Bk where(h,k) is the vertex.
The standard form is %28x+-h%29%5E2+=+4p%28y+-k%29, where the focus is (h,k + p)
the vertex form of a parabola opening right or left, x=a%28y-k%29%5E2+%2Bh where(h,k) is the vertex.
The standard form is %28y+-k%29%5E2+=+4p%28x+-h%29, where the focus is (h +p,k )