What is ((x-7)/4)^2+((y+3)/4)^2= 169/16 ?

The solution to ((x-7)/4)^2+((y+3)/4)^2= 169/16 is Circle with (a,b)=(7,-3),r=13

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((x-7)/4)^2+((y+3)/4)^2= 169/16

(\frac{x - 7}{4})^{2} + (\frac{y + 3}{4})^{2} = \frac{169}{16}

(a, b) = (7, - 3), r = 13

Solution steps

(\frac{x - 7}{4})^{2} + (\frac{y + 3}{4})^{2} = \frac{169}{16}

Rewrite

(\frac{x - 7}{4})^{2} + (\frac{y + 3}{4})^{2} = \frac{169}{16}

in the form of the standard circle equation

(x - 7)^{2} + (y - (- 3))^{2} = 1 3^{2}

Therefore the circle properties are:

(a, b) = (7, - 3), r = 13

x^{2} + y^{2} + 3 x + 4 y + 4 = 0

x^{2} + y^{2} - 10 x - 12 y + 45 = 0

(x - 4)^{2} + y^{2} = 16

x^{2} + y^{2} - 4 x - 12 y - 9 = 0

x^{2} + y^{2} - 14 x - 6 y + 48 = 0

What is ((x-7)/4)^2+((y+3)/4)^2= 169/16 ?
The solution to ((x-7)/4)^2+((y+3)/4)^2= 169/16 is Circle with (a,b)=(7,-3),r=13