Question

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

Answer 1

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 :)

Answer 2

Thank you so much!