What is (x+4)^2+(y-34)^2=1 ?

The solution to (x+4)^2+(y-34)^2=1 is Circle with (a,b)=(-4,34),r=1

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(x+4)^2+(y-34)^2=1

(x + 4)^{2} + (y - 34)^{2} = 1

(a, b) = (- 4, 34), r = 1

Solution steps

(x + 4)^{2} + (y - 34)^{2} = 1

Rewrite

(x + 4)^{2} + (y - 34)^{2} = 1

in the form of the standard circle equation

(x - (- 4))^{2} + (y - 34)^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (- 4, 34), r = 1

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

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

x^{2} + y^{2} - \frac{23}{2} x + \frac{11}{2} - \frac{13}{2} = 0

(x + 5)^{2} + (y - 8)^{2} = 25

x^{2} + y^{2} - 6 x - 8 y + 21 = 0

What is (x+4)^2+(y-34)^2=1 ?
The solution to (x+4)^2+(y-34)^2=1 is Circle with (a,b)=(-4,34),r=1