What is x^2+y^2-8x+5y-3=0 ?

The solution to x^2+y^2-8x+5y-3=0 is Circle with (a,b)=(4,-5/2),r=(sqrt(101))/2

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x^2+y^2-8x+5y-3=0

x^{2} + y^{2} - 8 x + 5 y - 3 = 0

(a, b) = (4, - \frac{5}{2}), r = \frac{101}{2}

Solution steps

x^{2} + y^{2} - 8 x + 5 y - 3 = 0

Rewrite

x^{2} + y^{2} - 8 x + 5 y - 3 = 0

in the form of the standard circle equation

(x - 4)^{2} + (y - (- \frac{5}{2}))^{2} = (\frac{101}{2})^{2}

Therefore the circle properties are:

(a, b) = (4, - \frac{5}{2}), r = \frac{101}{2}

4 x^{2} + 4 y^{2} - 10 x - 12 y = 20

16 x^{2} + 96 x + 16 y^{2} - 40 y - 315 + 315 = 0 + 315

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

x^{2} + (y - 5)^{2} = 100

\frac{( x - 1 ) ^{2}}{9} + \frac{( y + 4 ) ^{2}}{9} = 1

What is x^2+y^2-8x+5y-3=0 ?
The solution to x^2+y^2-8x+5y-3=0 is Circle with (a,b)=(4,-5/2),r=(sqrt(101))/2