How do you find the coordinates of the center of the circle x^2 +y^2 -4x+ 6x =12x2+y2−4x+6x=12?

1 Answer
ali ergin
Mar 11, 2016

r=7/2r=72

Explanation:

x^2+y^2-4x+6x=12x2+y2−4x+6x=12
x^2+y^2+2x-12=0x2+y2+2x−12=0
x^2+y^2+D x+E y+F=0x2+y2+Dx+Ey+F=0
C(a,b) " center coordinates"C(a,b) center coordinates
D=2" "E=0" "F=-12D=2 E=0 F=−12
a=-D/2=-2/2=-1a=−D2=−22=−1
b=-E/2=0/2=0b=−E2=02=0
r=1/2*sqrt(D^2+E^2-4*F)r=12⋅√D2+E2−4⋅F
r=1/2*sqrt((-1)^2+0+4*12)r=12⋅√(−1)2+0+4⋅12
r=1/2*sqrt(1+48)r=12⋅√1+48
r=1/2*sqrt 49r=12⋅√49
r=7/2r=72

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