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)=(7/4 ,-3/4),r= 13/4

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

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

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

(x - 2)^{2} + (y - 1)^{2} = 16

x^{2} + y^{2} = 72

x^{2} + y^{2} \leq 9

x^{2} + y^{2} - 4 x + 6 y - 3 = 0

x^{2} + y^{2} = 11

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)=(7/4 ,-3/4),r= 13/4