What is χ^2-14χ+γ^2+8γ+40=0 ?

The solution to χ^2-14χ+γ^2+8γ+40=0 is Circle with (a,b)=(7,-4),r=5

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χ^2-14χ+γ^2+8γ+40=0

χ^{2} - 14 χ + γ^{2} + 8 γ + 40 = 0

(a, b) = (7, - 4), r = 5

Solution steps

χ^{2} - 14 χ + γ^{2} + 8 γ + 40 = 0

Rewrite

χ^{2} - 14 χ + γ^{2} + 8 γ + 40 = 0

in the form of the standard circle equation

(χ - 7)^{2} + (γ - (- 4))^{2} = 5^{2}

Therefore the circle properties are:

(a, b) = (7, - 4), r = 5

x^{2} + y^{2} - 20 x + 8 y + 115 = 0

3 x^{2} + 3 y^{2} - 6 x + 9 y - 3 = 0

(x - 6)^{2} + (y - 5)^{2} = 16

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

(3 X - 2)^{2} + (3 Y + 6)^{2} = 144

What is χ^2-14χ+γ^2+8γ+40=0 ?
The solution to χ^2-14χ+γ^2+8γ+40=0 is Circle with (a,b)=(7,-4),r=5