What is (x+3)^2+(y-9)^2=21 ?

The solution to (x+3)^2+(y-9)^2=21 is Circle with (a,b)=(-3,9),r=sqrt(21)

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(x+3)^2+(y-9)^2=21

(x + 3)^{2} + (y - 9)^{2} = 21

(a, b) = (- 3, 9), r = 21

Solution steps

(x + 3)^{2} + (y - 9)^{2} = 21

Rewrite

(x + 3)^{2} + (y - 9)^{2} = 21

in the form of the standard circle equation

(x - (- 3))^{2} + (y - 9)^{2} = (21)^{2}

Therefore the circle properties are:

(a, b) = (- 3, 9), r = 21

5 x^{2} + 5 y^{2} + 10 x - y = 10

x^{2} + (y - 1)^{2} - 5 \leq 0

(x + 3)^{2} + (y + 5)^{2} = 5

(x + 3)^{2} + (y + 5)^{2} = 3

3 = - 6 x + 4 y + x^{2} + y^{2}

What is (x+3)^2+(y-9)^2=21 ?
The solution to (x+3)^2+(y-9)^2=21 is Circle with (a,b)=(-3,9),r=sqrt(21)