What is (x-4)^{(2)}+(y-7)^{(2)}=4 ?

The solution to (x-4)^{(2)}+(y-7)^{(2)}=4 is Circle with (a,b)=(4,7),r=2

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(x-4)^{(2)}+(y-7)^{(2)}=4

(x - 4)^{(2)} + (y - 7)^{(2)} = 4

(a, b) = (4, 7), r = 2

Solution steps

(x - 4)^{2} + (y - 7)^{2} = 4

Rewrite

(x - 4)^{2} + (y - 7)^{2} = 4

in the form of the standard circle equation

(x - 4)^{2} + (y - 7)^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (4, 7), r = 2

4 x^{2} - 8 \geq - 4 y^{2}

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

(a + b) 2 = a^{2} + b^{2}

x^{2} - 10 x + 25 + y^{2} - 8 y + 16 = 64

(x - 2)^{2} + (y - 4)^{2} = 25

What is (x-4)^{(2)}+(y-7)^{(2)}=4 ?
The solution to (x-4)^{(2)}+(y-7)^{(2)}=4 is Circle with (a,b)=(4,7),r=2