(A) r = 4 $\quad$ (B) r = 9 $\quad$ (C) r = 3 $\quad$ (D) r = 6

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Equation of circle $=x^2+y^2+2gx+2fy+c=0$

Now consider a circle having (h,k) as centre and radius a, Equation of this circle is $(x-h)^2+(y-k)^2=a^2$

$\Rightarrow$ Equation of a circle with it's center at (2,1), and radius $r$ is x^2+y^2-4x-2y+5-r^2=0$

The other given circle is $x^2+y^2-2x-6y+6=0$ (Note: it's center can be worked out to be $= (1,3)$

The equation of the common chord can be got by subtracting the above two equations, $\rightarrow$ $2x - 4y+1 = r^2$

This chord is the diameter of the second circle, which means that it's center (1,3) must be a solution of the above equation.

$\Rightarrow 2 \times 1 - 4 \times 3 + 1 = r^2 \rightarrow r^2 = 9 \rightarrow r = 3$

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