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Find the center and radius of the circle whose equation is 2x^2-3x+2y^2+4y-20=0

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2 Answers

 2 X2 -3X + 2y2 +4y -20 = 0
 
 2 ( x2 - 3/2X ) + 2 ( Y2 + 2Y ) = 20
 
2 ( X2 - 3/2 X + 9/16 ) + 2 ( Y2 + 2Y + 1 ) = 20 +9/8 + 2
 
2 ( X - 3/4)2 + 2 ( Y + 1 )  = 185/8
 
    ( X - 3/2)2 + ( Y + 1 )2 = 185/8
   
       Center ( 3/4, -1)   r = √(185/8) = 4.81
The equation of the circle has form:
 
(x-a)2+(y-b)2=R2, where point (a,b) is the center of the circle, R is its radius.
 
In your case,
2x2-3x+2y2+4y-20=0 Divide both sides by 2 to get:
 
x2-3/2*x+y2+2y-10=0; Let us work on y-coordinate of the center first.
 
y2+2y-10=y2+2y+1-11=(y2+2y+1)-11=(y+1)2-11. Now we made y-part in the form (y-b)2, here b=-1.
 
Back to the x-part;
 
x2-3x/2+(y+1)2-11=0; (x2-2*3/4*x+9/16)-9/16+(y+1)2-11=0;
 
Move -9/16 and -11 to the right side to get:
 
(x2-2*3x/4+9/16)+(y+1)2=11+9/16; ⇔ (x-3/4)2+(y+1)2=185/16;
 
This is now a canonical form of the equation of a circle. Its center is at the point (3/4;-1)