What is (x+4)^2+(y+12)^2-16=0 ?

The solution to (x+4)^2+(y+12)^2-16=0 is Circle with (a,b)=(-4,-12),r=4

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(x+4)^2+(y+12)^2-16=0

(x + 4)^{2} + (y + 12)^{2} - 16 = 0

(a, b) = (- 4, - 12), r = 4

Solution steps

(x + 4)^{2} + (y + 12)^{2} - 16 = 0

Rewrite

(x + 4)^{2} + (y + 12)^{2} - 16 = 0

in the form of the standard circle equation

(x - (- 4))^{2} + (y - (- 12))^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (- 4, - 12), r = 4

x^{2} + y^{2} + 6 x + 16 y - 8 = 0

4 x^{2} + 4 y^{2} + 8 x - 4 y - 3 = 0

(x - \frac{1}{2 ( \frac{3}{2} )})^{2} + y^{2} = (\frac{1}{2 ( \frac{3}{2} )})^{2}

x^{2} + y^{2} - 25 x = - 1 - 3 y

(x - 11)^{2} + (y - 8)^{2} = 268

What is (x+4)^2+(y+12)^2-16=0 ?
The solution to (x+4)^2+(y+12)^2-16=0 is Circle with (a,b)=(-4,-12),r=4