What is (x-3)^2+(y-5)^2=14^2 ?

The solution to (x-3)^2+(y-5)^2=14^2 is Circle with (a,b)=(3,5),r=14

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(x-3)^2+(y-5)^2=14^2

(x - 3)^{2} + (y - 5)^{2} = 1 4^{2}

(a, b) = (3, 5), r = 14

Solution steps

(x - 3)^{2} + (y - 5)^{2} = 1 4^{2}

Rewrite

(x - 3)^{2} + (y - 5)^{2} = 1 4^{2}

in the form of the standard circle equation

(x - 3)^{2} + (y - 5)^{2} = 1 4^{2}

Therefore the circle properties are:

(a, b) = (3, 5), r = 14

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

2 x^{2} + 3 x + 2 y^{2} + 7 y = 57.75

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

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

(x - 6)^{2} + (y + 5)^{2} = 4^{2}

What is (x-3)^2+(y-5)^2=14^2 ?
The solution to (x-3)^2+(y-5)^2=14^2 is Circle with (a,b)=(3,5),r=14