What is X^2+Y^2-5X+4Y-46=0 ?

The solution to X^2+Y^2-5X+4Y-46=0 is Circle with (a,b)=(5/2 ,-2),r= 15/2

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X^2+Y^2-5X+4Y-46=0

X^{2} + Y^{2} - 5 X + 4 Y - 46 = 0

(a, b) = (\frac{5}{2}, - 2), r = \frac{15}{2}

Solution steps

X^{2} + Y^{2} - 5 X + 4 Y - 46 = 0

Rewrite

X^{2} + Y^{2} - 5 X + 4 Y - 46 = 0

in the form of the standard circle equation

(X - \frac{5}{2})^{2} + (Y - (- 2))^{2} = (\frac{15}{2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{5}{2}, - 2), r = \frac{15}{2}

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

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

x^{2} + y^{2} - 10 x + 12 y = 3

x^{2} + y^{2} + 4 x + 2 y = 1

x^{2} + y^{2} - 22 x + 20 y + 196 = 0

What is X^2+Y^2-5X+4Y-46=0 ?
The solution to X^2+Y^2-5X+4Y-46=0 is Circle with (a,b)=(5/2 ,-2),r= 15/2