What is x^2+y^2= x/4-y/5 ?

The solution to x^2+y^2= x/4-y/5 is Circle with (a,b)=(1/8 ,-1/10),r=(sqrt(41))/(40)

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x^2+y^2= x/4-y/5

x^{2} + y^{2} = \frac{x}{4} - \frac{y}{5}

(a, b) = (\frac{1}{8}, - \frac{1}{10}), r = \frac{41}{40}

Solution steps

x^{2} + y^{2} = \frac{x}{4} - \frac{y}{5}

Rewrite

x^{2} + y^{2} = \frac{x}{4} - \frac{y}{5}

in the form of the standard circle equation

(x - \frac{1}{8})^{2} + (y - (- \frac{1}{10}))^{2} = (\frac{41}{40})^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{8}, - \frac{1}{10}), r = \frac{41}{40}

(x - 6)^{2} + (y - 2)^{2} = - 9^{2}

x^{2} + y^{2} - 8 x - 8 y + 23 = 0

(x + 2)^{2} + (y - 4)^{2} = 16

(x + 2)^{2} + (y - 4)^{2} = 11

y^{2} = 2 + x - x^{2}

What is x^2+y^2= x/4-y/5 ?
The solution to x^2+y^2= x/4-y/5 is Circle with (a,b)=(1/8 ,-1/10),r=(sqrt(41))/(40)