If the line 3x + 4y - 1 = 0 touches the circl | vedclass

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(A) The equation of the circle is $(x - 1)^2 + (y - 2)^2 = r^2$,which has its center at $(1, 2)$ and radius $r$.If a line $Ax + By + C = 0$ touches a

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$2$

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(A) The equation of the circle is $(x - 1)^2 + (y - 2)^2 = r^2$,which has its center at $(1, 2)$ and radius $r$.If a line $Ax + By + C = 0$ touches a circle,the perpendicular distance from the center of the circle to the line must be equal to the radius $r$.The perpendicular distance $d$ from the point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.Substituting the values,we get $r = \frac{|3(1) + 4(2) - 1|}{\sqrt{3^2 + 4^2}}$.$r = \frac{|3 + 8 - 1|}{\sqrt{9 + 16}} = \frac{|10|}{\sqrt{25}} = \frac{10}{5} = 2$.Thus,the value of $r$ is $2$.

If the line $3x + 4y - 1 = 0$ touches the circle $(x - 1)^2 + (y - 2)^2 = r^2$,then the value of $r$ will be

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