What is (x-8)^2+(y-6)^2=1^2 ?

The solution to (x-8)^2+(y-6)^2=1^2 is Circle with (a,b)=(8,6),r=1

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

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

(a, b) = (8, 6), r = 1

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (8, 6), r = 1

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

(x - 8)^{2} + (y - 3)^{2} = 3

What is (x-8)^2+(y-6)^2=1^2 ?
The solution to (x-8)^2+(y-6)^2=1^2 is Circle with (a,b)=(8,6),r=1