What is 0=y^2-4y+(x+4)^2 ?

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

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0=y^2-4y+(x+4)^2

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

(a, b) = (- 4, 2), r = 2

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 4, 2), r = 2

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

(x - 2)^{2} + (y - (- 4))^{2} = 1. 7^{2}

x^{2} + y^{2} + y - 11 = 0

x^{2} - 154 x + y^{2} - 160 y = 0

\frac{( x - 1 ) ^{2}}{16} + \frac{( y + 2 ) ^{2}}{16} = 10

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