What is x^2+y^2-9x+4y-98=0 ?

The solution to x^2+y^2-9x+4y-98=0 is Circle with (a,b)=(9/2 ,-2),r=(sqrt(489))/2

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x^2+y^2-9x+4y-98=0

x^{2} + y^{2} - 9 x + 4 y - 98 = 0

(a, b) = (\frac{9}{2}, - 2), r = \frac{489}{2}

Solution steps

x^{2} + y^{2} - 9 x + 4 y - 98 = 0

Rewrite

x^{2} + y^{2} - 9 x + 4 y - 98 = 0

in the form of the standard circle equation

(x - \frac{9}{2})^{2} + (y - (- 2))^{2} = (\frac{489}{2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{9}{2}, - 2), r = \frac{489}{2}

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x^{2} + y^{2} \leq 13

9 x^{2} + 9 y^{2} + 54 x - 36 y + 17 = 0

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

What is x^2+y^2-9x+4y-98=0 ?
The solution to x^2+y^2-9x+4y-98=0 is Circle with (a,b)=(9/2 ,-2),r=(sqrt(489))/2