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)   (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
|C(2,0 ) and r = 6
Note:
Standard Form of an Equation of a Circle is
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
where Pt(h,k) is the center and r is the radius
Standard Form of an Equation of an Ellipse is 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 ±are the foci distances from center: a > b
Standard Form of an Equation of an Hyperbola opening right and left is:
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:
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, where(h,k) is the vertex.
The standard form is , where the focus is (h,k + p)
the vertex form of a parabola opening right or left, where(h,k) is the vertex.
The standard form is , where the focus is (h +p,k )
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