What is the radius x^2+y^2+3x-5y+1=0 ?

The radius x^2+y^2+3x-5y+1=0 is r=sqrt(15/2)

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radius x^2+y^2+3x-5y+1=0

radius

x^{2} + y^{2} + 3 x - 5 y + 1 = 0

r = \frac{15}{2}

Solution steps

x^{2} + y^{2} + 3 x - 5 y + 1 = 0

Rewrite

x^{2} + y^{2} + 3 x - 5 y + 1 = 0

in the form of the standard circle equation

(x - (- \frac{3}{2}))^{2} + (y - \frac{5}{2})^{2} = (\frac{15}{2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{3}{2}, \frac{5}{2}), r = \frac{15}{2}

And the radius is:

r = \frac{15}{2}

x^{2} + y^{2} = 1 + 2 y

x^{2} + y^{2} - 8 \cdot + 7 = 0

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

x^{2} + y^{2} + 7 x - 4 y - 2 = 0

y^{2} + x^{2} = 25

What is the radius x^2+y^2+3x-5y+1=0 ?
The radius x^2+y^2+3x-5y+1=0 is r=sqrt(15/2)