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

The center ((x-7)/4)^2+((y+3)/4)^2= 169/16 is (7,-3)

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

center

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

(7, - 3)

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

And the center is:

(7, - 3)

x^{2} + y^{2} + 4 x + 6 y + 9 = 0

x^{2} - 14 x + 49 + y^{2} + 8 y + 16 - 25 = 0

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

x^{2} + y^{2} + 2 x + 10 y + 18 = 0

x^{2} + (y - 1)^{2} = 9

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