What is (x+4)^2+(y+1)^2=100 ?

The solution to (x+4)^2+(y+1)^2=100 is Circle with (a,b)=(-4,-1),r=10

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(x+4)^2+(y+1)^2=100

(x + 4)^{2} + (y + 1)^{2} = 100

(a, b) = (- 4, - 1), r = 10

Solution steps

(x + 4)^{2} + (y + 1)^{2} = 100

Rewrite

(x + 4)^{2} + (y + 1)^{2} = 100

in the form of the standard circle equation

(x - (- 4))^{2} + (y - (- 1))^{2} = 1 0^{2}

Therefore the circle properties are:

(a, b) = (- 4, - 1), r = 10

(x - 8)^{2} + (y - 6)^{2} = 1^{2}

x^{2} + (y - \frac{25}{3})^{2} = 144

\frac{x ^{2}}{15} + \frac{y ^{2}}{15} - \frac{2 x}{15} - \frac{6}{15} = 0

(x - 7)^{2} + (y + 13)^{2} = 81

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

What is (x+4)^2+(y+1)^2=100 ?
The solution to (x+4)^2+(y+1)^2=100 is Circle with (a,b)=(-4,-1),r=10