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Q)

The radius of general equation of circle is

$\begin{array}{1 1}(a)\;a=\sqrt{g^2+f^2-c}\\(b)\;a=\sqrt{g^2-f^2+c}\\(c)\;a=\sqrt{g^2+f^2+c}\\(d)\;\text{None of these}\end{array}$

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A)
Equation of circle $=x^2+y^2+2gx+2fy+c=0$-------(1)
Now consider a circle having (h,k) as centre and radius a
Equation of this circle is $(x-h)^2+(y-k)^2=a^2$
(i.e) $k^2+y^2-2hx-2yk+h^2+k^2-a^2=0$--------(2)
Comparing (1) & (2)
$h=-g,k=-f,a=\sqrt{g^2+f^2-c}$
Hence radius a=$\sqrt{g^2+f^2-c}$
Hence (a) is the correct answer.
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