What is the radius 3.5(x+2.2)^2+3.5(y-11.1)^2-21=0 ?

The radius 3.5(x+2.2)^2+3.5(y-11.1)^2-21=0 is r=sqrt(6)

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radius 3.5(x+2.2)^2+3.5(y-11.1)^2-21=0

radius

3.5 (x + 2.2)^{2} + 3.5 (y - 11.1)^{2} - 21 = 0

r = 6

Solution steps

3.5 (x + 2.2)^{2} + 3.5 (y - 11.1)^{2} - 21 = 0

Rewrite

3.5 (x + 2.2)^{2} + 3.5 (y - 11.1)^{2} - 21 = 0

in the form of the standard circle equation

(x - (- 2.2))^{2} + (y - 11.1)^{2} = (6)^{2}

Therefore the circle properties are:

(a, b) = (- 2.2, 11.1), r = 6

And the radius is:

r = 6

x^{2} + y^{2} + 6 x - 2 y + 1 = 0

x^{2} + y^{2} + 10 x + 12 y + 25 = 0

x^{2} + y^{2} - 10 x - 8 y = - 5

x^{2} - 2 x + y^{2} - 8 y + 1 = 0

x^{2} + y^{2} - 8 x + 12 y + 10 = 0

What is the radius 3.5(x+2.2)^2+3.5(y-11.1)^2-21=0 ?
The radius 3.5(x+2.2)^2+3.5(y-11.1)^2-21=0 is r=sqrt(6)