What is a^2-28+b^2=7 ?

The solution to a^2-28+b^2=7 is Circle with (a,b)=(0,0),r=sqrt(35)

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a^2-28+b^2=7

a^{2} - 28 + b^{2} = 7

(a, b) = (0, 0), r = 35

Solution steps

a^{2} - 28 + b^{2} = 7

Rewrite

a^{2} - 28 + b^{2} = 7

in the form of the standard circle equation

(a - 0)^{2} + (b - 0)^{2} = (35)^{2}

Therefore the circle properties are:

(a, b) = (0, 0), r = 35

\frac{( x - 5 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{4} = 1

(x - 6)^{2} + (y - 4)^{2} = 8^{2}

(x - 4)^{2} + (k (x) - 3)^{2} = 5^{2}

x^{2} + y^{2} + 6 y - 18 = 0

3 x^{2} + 3 y^{2} + 5 x - 6 y + 1 = 0

What is a^2-28+b^2=7 ?
The solution to a^2-28+b^2=7 is Circle with (a,b)=(0,0),r=sqrt(35)