Given Equation :
x2+y2−2x+4y+1=0x^2+y^2-2x+4y+1=0x2+y2−2x+4y+1=0
which is the form of general equation of circle :
x2+y2+2gx+2fy+c=0x^2+y^2+2gx+2fy+c=0x2+y2+2gx+2fy+c=0
with center at (−g,−f)(-g,-f)(−g,−f) and radius being r=g2+f2−c\ r=\sqrt{g^2+f^2-c} r=g2+f2−c
In given equation,
g=−22=−1 ⟹ −g=1g=\frac{-2}{2}=-1\implies -g=1g=2−2=−1⟹−g=1
And f=42=2 ⟹ −f=−2f=\frac{4}{2}=2\implies -f=-2f=24=2⟹−f=−2
And c=1c=1c=1
So, center of circle is (−g,−f)≡(1,−2)(-g,-f)\equiv(1,-2)(−g,−f)≡(1,−2)
And radius of circle will be r=g2+f2−c=(−1)2+22−1=4=2\ r=\sqrt{g^2+f^2-c}=\sqrt{(-1)^2+2^2-1}=\sqrt{4}=2 r=g2+f2−c=(−1)2+22−1=4=2 units.
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