Let C_1 be the circle of radius 1 with center | vedclass

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(A) The circle $C_1$ is $x^2+y^2=1$ and $C_2$ is $(x-4)^2+(y-1)^2=r^2$.The radical axis of two circles is given by $S_1 - S_2 = 0$.$x^2+y^2-1 - ((x-4)

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$2$

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(A) The circle $C_1$ is $x^2+y^2=1$ and $C_2$ is $(x-4)^2+(y-1)^2=r^2$.The radical axis of two circles is given by $S_1 - S_2 = 0$.$x^2+y^2-1 - ((x-4)^2+(y-1)^2-r^2) = 0$$x^2+y^2-1 - (x^2-8x+16+y^2-2y+1-r^2) = 0$$8x+2y-18+r^2 = 0$,which simplifies to $8x+2y = 18-r^2$.The line joining the midpoints of the common tangents is the radical axis of the two circles.This line meets the $x$-axis at $B$. Setting $y=0$ in the radical axis equation,we get $8x = 18-r^2$,so $x = \frac{18-r^2}{8}$.Thus,$B = \left(\frac{18-r^2}{8}, 0\right)$.Given $A=(4,1)$ and $AB=\sqrt{5}$,we have $AB^2 = 5$.$\left(\frac{18-r^2}{8}-4\right)^2 + (0-1)^2 = 5$$\left(\frac{18-r^2-32}{8}\right)^2 + 1 = 5$$\left(\frac{-(14+r^2)}{8}\right)^2 = 4$$\frac{14+r^2}{8} = 2$ (since $r^2 > 0$,the expression inside the square must be positive for the root to be $2$)$14+r^2 = 16$,which gives $r^2 = 2$.

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