What is x^2+y^2+7x-8y=3 ?

The solution to x^2+y^2+7x-8y=3 is Circle with (a,b)=(-7/2 ,4),r=(5sqrt(5))/2

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x^2+y^2+7x-8y=3

x^{2} + y^{2} + 7 x - 8 y = 3

(a, b) = (- \frac{7}{2}, 4), r = \frac{5 5}{2}

Solution steps

x^{2} + y^{2} + 7 x - 8 y = 3

Rewrite

x^{2} + y^{2} + 7 x - 8 y = 3

in the form of the standard circle equation

(x - (- \frac{7}{2}))^{2} + (y - 4)^{2} = (\frac{5 5}{2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{7}{2}, 4), r = \frac{5 5}{2}

x^{2} + y^{2} + 6 x - 8 y = 0

y^{2} = 49 - x^{2}

x^{2} + y^{2} + 16 x - 8 y + 76 = 0

x^{2} + (y - 1)^{2} = 49

x^{2} + y^{2} + 20 x + 20 y + 164 = 0

What is x^2+y^2+7x-8y=3 ?
The solution to x^2+y^2+7x-8y=3 is Circle with (a,b)=(-7/2 ,4),r=(5sqrt(5))/2