What is the center x^2+y^2-2/3 x-5/3 y-5/9 =0 ?

The center x^2+y^2-2/3 x-5/3 y-5/9 =0 is (1/3 , 5/6)

Popular >

center x^2+y^2-2/3 x-5/3 y-5/9 =0

center

x^{2} + y^{2} - \frac{2}{3} x - \frac{5}{3} y - \frac{5}{9} = 0

(\frac{1}{3}, \frac{5}{6})

Solution steps

x^{2} + y^{2} - \frac{2}{3} x - \frac{5}{3} y - \frac{5}{9} = 0

Rewrite

x^{2} + y^{2} - \frac{2}{3} x - \frac{5}{3} y - \frac{5}{9} = 0

in the form of the standard circle equation

(x - \frac{1}{3})^{2} + (y - \frac{5}{6})^{2} = (\frac{7}{6})^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{3}, \frac{5}{6}), r = \frac{7}{6}

And the center is:

(\frac{1}{3}, \frac{5}{6})

\frac{( x + 2 ) ^{2}}{16} + \frac{( y + 5 ) ^{2}}{16} = 1

9 x^{2} - 42 x + 9 y^{2} + 84 y + 65 = 0

x^{2} + y^{2} - 8 y + 7 = 0

8 x^{2} + 8 y^{2} - 20 x + 64 y - 371 = 0

(y + 8)^{2} + (x + 2)^{2} = 72

What is the center x^2+y^2-2/3 x-5/3 y-5/9 =0 ?
The center x^2+y^2-2/3 x-5/3 y-5/9 =0 is (1/3 , 5/6)