What is (8x^2-12x+8y^2+20y+17)/8 =2 ?

The solution to (8x^2-12x+8y^2+20y+17)/8 =2 is Circle with (a,b)=(3/4 ,-5/4),r=sqrt(2)

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(8x^2-12x+8y^2+20y+17)/8 =2

\frac{8 x ^{2} - 12 x + 8 y ^{2} + 20 y + 17}{8} = 2

(a, b) = (\frac{3}{4}, - \frac{5}{4}), r = 2

Solution steps

\frac{8 x ^{2} - 12 x + 8 y ^{2} + 20 y + 17}{8} = 2

Rewrite

\frac{8 x ^{2} - 12 x + 8 y ^{2} + 20 y + 17}{8} = 2

in the form of the standard circle equation

(x - \frac{3}{4})^{2} + (y - (- \frac{5}{4}))^{2} = (2)^{2}

Therefore the circle properties are:

(a, b) = (\frac{3}{4}, - \frac{5}{4}), r = 2

(x + 1)^{2} + (y - 4)^{2} = 16

x^{2} + (y - 3)^{2} = 100

(x - 15)^{2} + (y + 15)^{2} = 256

[5 (x + 4)]^{2} + 25 (y - 2)^{2} = 625

x^{2} - 1 = 0 - y^{2} + 1

What is (8x^2-12x+8y^2+20y+17)/8 =2 ?
The solution to (8x^2-12x+8y^2+20y+17)/8 =2 is Circle with (a,b)=(3/4 ,-5/4),r=sqrt(2)