What is (x-3)^2+(y+4)^2=5 ?

The solution to (x-3)^2+(y+4)^2=5 is Circle with (a,b)=(3,-4),r=sqrt(5)

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(x-3)^2+(y+4)^2=5

(x - 3)^{2} + (y + 4)^{2} = 5

(a, b) = (3, - 4), r = 5

Solution steps

(x - 3)^{2} + (y + 4)^{2} = 5

Rewrite

(x - 3)^{2} + (y + 4)^{2} = 5

in the form of the standard circle equation

(x - 3)^{2} + (y - (- 4))^{2} = (5)^{2}

Therefore the circle properties are:

(a, b) = (3, - 4), r = 5

(x - 3)^{2} + (y + 4)^{2} = 3

4 x^{2} + 12 x + 4 y^{2} + 16 - 11 = 0

x^{2} = 50 - y^{2}

215 = a^{2} + b^{2}

(x)^{2} + (y - 4)^{2} = 73

What is (x-3)^2+(y+4)^2=5 ?
The solution to (x-3)^2+(y+4)^2=5 is Circle with (a,b)=(3,-4),r=sqrt(5)