What is 4x^2+4y^2-16x+40y-7=0 ?

The solution to 4x^2+4y^2-16x+40y-7=0 is Circle with (a,b)=(2,-5),r=(sqrt(123))/2

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4x^2+4y^2-16x+40y-7=0

4 x^{2} + 4 y^{2} - 16 x + 40 y - 7 = 0

(a, b) = (2, - 5), r = \frac{123}{2}

Solution steps

4 x^{2} + 4 y^{2} - 16 x + 40 y - 7 = 0

Rewrite

4 x^{2} + 4 y^{2} - 16 x + 40 y - 7 = 0

in the form of the standard circle equation

(x - 2)^{2} + (y - (- 5))^{2} = (\frac{123}{2})^{2}

Therefore the circle properties are:

(a, b) = (2, - 5), r = \frac{123}{2}

(x - 11)^{2} + (y - 12)^{2} = 25

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

x^{2} + y^{2} - 6 x - 10 y + 2 = 0

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

x^{2} + y^{2} - 8 x + 2 y - 24 = 0

What is 4x^2+4y^2-16x+40y-7=0 ?
The solution to 4x^2+4y^2-16x+40y-7=0 is Circle with (a,b)=(2,-5),r=(sqrt(123))/2