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 (7/4 ,-3/4)

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

center

(\frac{7}{4}, - \frac{3}{4})

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - \frac{7}{4})^{2} + (y - (- \frac{3}{4}))^{2} = (\frac{13}{4})^{2}

Therefore the circle properties are:

(a, b) = (\frac{7}{4}, - \frac{3}{4}), r = \frac{13}{4}

And the center is:

(\frac{7}{4}, - \frac{3}{4})

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 (7/4 ,-3/4)