The center of the circle is (?,?)

The radius of the circle is ?

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The center of the circle is (?,?)

The radius of the circle is ?

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2 X^{2} -3X + 2y^{2} +4y -20 = 0

2 ( x^{2 }- 3/2X ) + 2 ( Y^{2} + 2Y ) = 20

2 ( X^{2} - 3/2 X + 9/16 ) + 2 ( Y^{2} + 2Y + 1 ) = 20 +9/8 + 2

2 ( X - 3/4)^{2} + 2 ( Y + 1 ) ^{2 } = 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}=R^{2}, where point (a,b) is the center of the circle, R is its radius.

In your case,

2x^{2}-3x+2y^{2}+4y-20=0 Divide both sides by 2 to get:

x^{2}-3/2*x+y^{2}+2y-10=0; Let us work on y-coordinate of the center first.

y^{2}+2y-10=y^{2}+2y+1-11=(y^{2}+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;

x^{2}-3x/2+(y+1)^{2}-11=0; (x^{2}-2*3/4*x+9/16)-9/16+(y+1)^{2}-11=0;

Move -9/16 and -11 to the right side to get:

(x^{2}-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)**

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