A line meets the coordinate axes at A and B. | vedclass

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(C) Let the line $\frac{x}{a} + \frac{y}{b} = 1$ meet the coordinate axes at $A(a, 0)$ and $B(0, b)$.Since $	riangle OAB$ is a right-angled triangle,

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$d_1 + d_2$

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(C) Let the line $\frac{x}{a} + \frac{y}{b} = 1$ meet the coordinate axes at $A(a, 0)$ and $B(0, b)$.Since $	riangle OAB$ is a right-angled triangle,the circumcircle has the hypotenuse $AB$ as its diameter.The center of the circle is $C = (\frac{a}{2}, \frac{b}{2})$ and the radius is $r = \frac{\sqrt{a^2 + b^2}}{2}$.The slope of $OC$ is $\frac{b}{a}$,so the slope of the tangent at the origin $O$ is $-\frac{a}{b}$.The equation of the tangent at $O$ is $ax + by = 0$.The distance $d_1$ from $A(a, 0)$ to the tangent $ax + by = 0$ is $d_1 = \frac{|a(a) + b(0)|}{\sqrt{a^2 + b^2}} = \frac{a^2}{\sqrt{a^2 + b^2}}$.The distance $d_2$ from $B(0, b)$ to the tangent $ax + by = 0$ is $d_2 = \frac{|a(0) + b(b)|}{\sqrt{a^2 + b^2}} = \frac{b^2}{\sqrt{a^2 + b^2}}$.Adding these distances,we get $d_1 + d_2 = \frac{a^2 + b^2}{\sqrt{a^2 + b^2}} = \sqrt{a^2 + b^2}$.Since the diameter is $2r = \sqrt{a^2 + b^2}$,the diameter is $d_1 + d_2$.

$A$ line meets the coordinate axes at $A$ and $B$. $A$ circle is circumscribed about the triangle $OAB$. If $d_1$ and $d_2$ are the distances of the tangent to the circle at the origin $O$ from the points $A$ and $B$ respectively,then the diameter of the circle is:

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