What is (x+1)^2+(y-1)^2=9(x-2)^2+9(y+1)^2 ?

The solution to (x+1)^2+(y-1)^2=9(x-2)^2+9(y+1)^2 is Circle with (a,b)=(19/8 ,-5/4),r=(3sqrt(13))/8

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(x+1)^2+(y-1)^2=9(x-2)^2+9(y+1)^2

(x + 1)^{2} + (y - 1)^{2} = 9 (x - 2)^{2} + 9 (y + 1)^{2}

(a, b) = (\frac{19}{8}, - \frac{5}{4}), r = \frac{3 13}{8}

Solution steps

(x + 1)^{2} + (y - 1)^{2} = 9 (x - 2)^{2} + 9 (y + 1)^{2}

Rewrite

(x + 1)^{2} + (y - 1)^{2} = 9 (x - 2)^{2} + 9 (y + 1)^{2}

in the form of the standard circle equation

(x - \frac{19}{8})^{2} + (y - (- \frac{5}{4}))^{2} = (\frac{3 13}{8})^{2}

Therefore the circle properties are:

(a, b) = (\frac{19}{8}, - \frac{5}{4}), r = \frac{3 13}{8}

(x - 9)^{2} + (y - 8)^{2} = 81

x^{2} + y^{2} - 16 x + 8 y + 55 = 0

x^{2} + y^{2} - 10 x + 14 y - 7 = 0

x^{2} + y^{2} = 47 + 3 x

x^{2} + y^{2} - 6 y = 55

What is (x+1)^2+(y-1)^2=9(x-2)^2+9(y+1)^2 ?
The solution to (x+1)^2+(y-1)^2=9(x-2)^2+9(y+1)^2 is Circle with (a,b)=(19/8 ,-5/4),r=(3sqrt(13))/8