What is (x-3)^2+(y+5)^2=100 ?

The solution to (x-3)^2+(y+5)^2=100 is Circle with (a,b)=(3,-5),r=10

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(x-3)^2+(y+5)^2=100

(x - 3)^{2} + (y + 5)^{2} = 100

(a, b) = (3, - 5), r = 10

Solution steps

(x - 3)^{2} + (y + 5)^{2} = 100

Rewrite

(x - 3)^{2} + (y + 5)^{2} = 100

in the form of the standard circle equation

(x - 3)^{2} + (y - (- 5))^{2} = 1 0^{2}

Therefore the circle properties are:

(a, b) = (3, - 5), r = 10

π x^{(2)} + π y^{(2)} + 8 x - 10 π y - 400 = 0

x^{2} + y^{2} + 14 x + 8 y + 3 = 0

x^{2} + y^{2} + 18 x - 30 y + 297 = 0

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

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

What is (x-3)^2+(y+5)^2=100 ?
The solution to (x-3)^2+(y+5)^2=100 is Circle with (a,b)=(3,-5),r=10