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(A) The radius of a circle that touches a line is equal to the perpendicular distance from the centre of the circle to that line.The formula for the p

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$2$

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(A) The radius of a circle that touches a line is equal to the perpendicular distance from the centre of the circle to that line.The formula for the perpendicular distance from a point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.Given the centre $(x_1, y_1) = (1, -3)$ and the line $3x - 4y - 5 = 0$,we have $A = 3$,$B = -4$,and $C = -5$.Substituting these values into the formula:$r = \frac{|3(1) - 4(-3) - 5|}{\sqrt{3^2 + (-4)^2}}$$r = \frac{|3 + 12 - 5|}{\sqrt{9 + 16}}$$r = \frac{|10|}{\sqrt{25}}$$r = \frac{10}{5} = 2$.Thus,the radius of the circle is $2$.

If a circle whose centre is $(1, -3)$ touches the line $3x - 4y - 5 = 0$,then the radius of the circle is

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