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(B) Let the equation of the line $AB$ be $\frac{x}{a} + \frac{y}{b} = 1$. The points are $A(a, 0)$ and $B(0, b)$.The circle passing through $O(0, 0)$,

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$m+n$

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(B) Let the equation of the line $AB$ be $\frac{x}{a} + \frac{y}{b} = 1$. The points are $A(a, 0)$ and $B(0, b)$.The circle passing through $O(0, 0)$, $A(a, 0)$, and $B(0, b)$ has the equation $x^2 + y^2 - ax - by = 0$.The tangent to the circle at the origin $(0, 0)$ is obtained by replacing $x^2$ with $0$, $y^2$ with $0$, $x$ with $\frac{x}{2}$, and $y$ with $\frac{y}{2}$.Thus, the tangent at the origin is $-ax - by = 0$, or $ax + by = 0$.The 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 distance $n$ from $B(0, b)$ to the line $ax + by = 0$ is $n = \frac{|a(0) + b(b)|}{\sqrt{a^2 + b^2}} = \frac{b^2}{\sqrt{a^2 + b^2}}$.The diameter of the circle is the length of the hypotenuse $AB = \sqrt{a^2 + b^2}$.From the expressions for $m$ and $n$, we have $m+n = \frac{a^2 + b^2}{\sqrt{a^2 + b^2}} = \sqrt{a^2 + b^2}$.Therefore, the diameter of the circle is $m+n$.

$A$ straight line meets the coordinate axes at $A$ and $B$. $A$ circle is circumscribed about the triangle $OAB$, where $O$ is the origin. If $m$ and $n$ are the distances of the tangent to the circle at the origin from the points $A$ and $B$ respectively, then the diameter of the circle is

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