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Mathematics 33 Online
OpenStudy (anonymous):

Find the radius and the coordinates of the center of the circle X^2+Y^2+(6/5)X-(8/5)Y-8=0

OpenStudy (amistre64):

Equation of a circle: (x+h)^2 + (y+k)^2 = r^2. (x^2 +(6/5)x + a) + (y^2 -(8/5)y +b) = r^2; where (a+b)-r^2 = -8. a=[(6/5)/2]^2 = (6/10)^2 = 36/100. b=[(8/5)/2]^2 = (8/10)^2 = 64/100 a+b= 100/100= 1 1-r^2=-8 or 1+8 = r^2 -> r^2=9. radius=sqrt(9)=3. Now, (x^2 +(6/5)x + 36/100) = [x+(6/10)]^2. Center X coordinate = (-6/10). And, (y^2 -(8/5)y +64/100) = [y-(8/10)]^2. Center Y coordinate = (8/10) That should be correct :)

OpenStudy (anonymous):

Thank you so much!

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