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(C) Let the coordinates of $A$ be $(a, 0)$ and $B$ be $(0, b)$. The circumcircle of $	riangle OAB$ passes through $(0, 0)$,$(a, 0)$,and $(0, b)$.The

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$m+n$

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(C) Let the coordinates of $A$ be $(a, 0)$ and $B$ be $(0, b)$. The circumcircle of $	riangle OAB$ passes through $(0, 0)$,$(a, 0)$,and $(0, b)$.The equation of the circle is $x^2 + y^2 - ax - by = 0$.The tangent to this circle at the origin $(0, 0)$ is obtained by setting the first-degree terms to zero: $-ax - by = 0$,or $ax + by = 0$.The perpendicular distance $m$ from $A(a, 0)$ to the line $ax + by = 0$ is $m = \frac{|a(a) + b(0)|}{\sqrt{a^2 + b^2}} = \frac{a^2}{\sqrt{a^2 + b^2}}$.The perpendicular distance $n$ from $B(0, b)$ to the line $ax + by = 0$ is $n = \frac{|a(0) + b(b)|}{\sqrt{a^2 + b^2}} = \frac{b^2}{\sqrt{a^2 + b^2}}$.Adding these distances,we get $m + n = \frac{a^2 + b^2}{\sqrt{a^2 + b^2}} = \sqrt{a^2 + b^2}$.The diameter of the circle $x^2 + y^2 - ax - by = 0$ is given by $\sqrt{g^2 + f^2 - c} 	imes 2 = \sqrt{(-a/2)^2 + (-b/2)^2} 	imes 2 = \sqrt{\frac{a^2+b^2}{4}} 	imes 2 = \sqrt{a^2 + b^2}$.Thus,the diameter is $m + n$.

$A$ line meets the coordinate axes at $A(a, 0)$ and $B(0, b)$. If the perpendicular distances from $A$ and $B$ to the tangent drawn at the origin to the circumcircle of $\triangle OAB$ are $m$ and $n$ respectively,then the diameter of that circle is

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