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} + y^{2} + 6 x - 22 y + 120 = 0

(x + 4)^{2} + (y + 5)^{2} = 36

(x - 6)^{2} + (y - 7)^{2} = 8^{2}

(x - 3)^{2} + (y - 2)^{2} = 5

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

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