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)