If the tangent to the circle {x^2} + {y^2} = | vedclass

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Question

(a) Obviously $r = \sqrt {{a^2} + {b^2}} $

Equation of $AB$ is $ax + by = {r^2}$ or $\frac{x}{{{r^2}/a}} + \frac{y}{{{r^2}/b}} = 1$

$ \Rightarro

Vedclass · Accepted Answer

$\frac{{{r^4}}}{{2ab}}$

Vedclass · Answer

(a) Obviously $r = \sqrt {{a^2} + {b^2}} $

Equation of $AB$ is $ax + by = {r^2}$ or $\frac{x}{{{r^2}/a}} + \frac{y}{{{r^2}/b}} = 1$

$ \Rightarrow OA = \frac{{{r^2}}}{a}$ and $OB = \frac{{{r^2}}}{b}$

Hence the area is $\frac{1}{2}.\frac{{{r^2}}}{a}.\frac{{{r^2}}}{b} $

$= \frac{1}{2}\frac{{{r^4}}}{{ab}}$.

If the tangent to the circle ${x^2} + {y^2} = {r^2}$ at the point $(a, b)$ meets the coordinate axes at the point $A$ and $B$, and $O$ is the origin, then the area of the triangle $OAB$ is

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