What is (-2/5-x)^2+(-6/5-y)^2=10 ?

The solution to (-2/5-x)^2+(-6/5-y)^2=10 is Circle with (a,b)=(-2/5 ,-6/5),r=sqrt(10)

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(-2/5-x)^2+(-6/5-y)^2=10

(- \frac{2}{5} - x)^{2} + (- \frac{6}{5} - y)^{2} = 10

(a, b) = (- \frac{2}{5}, - \frac{6}{5}), r = 10

Solution steps

(- \frac{2}{5} - x)^{2} + (- \frac{6}{5} - y)^{2} = 10

Rewrite

(- \frac{2}{5} - x)^{2} + (- \frac{6}{5} - y)^{2} = 10

in the form of the standard circle equation

(x - (- \frac{2}{5}))^{2} + (y - (- \frac{6}{5}))^{2} = (10)^{2}

Therefore the circle properties are:

(a, b) = (- \frac{2}{5}, - \frac{6}{5}), r = 10

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(x - 1.5)^{2} + (y - 1)^{2} = 3

x^{2} + y^{2} - 6 x + 8 y - 12 = 0

5 (x^{2} - 1.8) + 5 (y^{2} - 28) + 26 = 0

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What is (-2/5-x)^2+(-6/5-y)^2=10 ?
The solution to (-2/5-x)^2+(-6/5-y)^2=10 is Circle with (a,b)=(-2/5 ,-6/5),r=sqrt(10)