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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

(x + 7)^{2} + (y - 8)^{2} = 27

x^{2} + y^{2} - 6 x + 1 = 0

4 x^{2} + 4 y^{2} - 8 x + 16 y + 4 = 0

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

x^{2} + y^{2} + 2 (\frac{29}{22}) x + 2 (\frac{29}{22}) y + (- \frac{142}{11}) = 0

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