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A line meets the coordinate axes at $A$ and $B$. If the perpendicular distances from $A$ and $B$ to the tangent drawn at the origin to the circumcircle of $\triangle O A B$ are $m$ and $n$ respectively, then the diameter of that circle Is
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Verified Answer
The correct answer is:
$m+n$
Equation of circle passing through
$(a, 0)(0, b)$ and $(0,0)$ is $x^2+y^2-a x-b y=0$
Equation of tangent at origin, $-a x-b y=0$
$$
\begin{aligned}
& \Rightarrow \quad a x+b y=0 \\
& P A=\frac{a^2}{\sqrt{a^2+b^2}} \\
& \therefore \quad m=\left|\frac{a^2}{\sqrt{a^2+b^2}}\right| \\
& \text { and } B Q=n=\left|\frac{b^2}{\sqrt{a^2+b^2}}\right| \\
& m^2+n^2=\frac{a^4+b^4}{a^2+b^2} \\
& m n=\frac{a^2 b^2}{a^2+b^2} \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \because \quad m^2+n^2+2 m n=\frac{a^4+b^4+2 a^2 b^2}{a^2+b^2} \\
& \qquad \begin{aligned}
(m+n)^2=\frac{\left(a^2+b^2\right)^2}{a^2+b^2}=a^2+b^2
\end{aligned} \\
& \therefore \text { Diametre of circle }=\sqrt{a^2+b^2}=\sqrt{(m+n)^2} \\
& =m+n
\end{aligned}
$$
$(a, 0)(0, b)$ and $(0,0)$ is $x^2+y^2-a x-b y=0$
Equation of tangent at origin, $-a x-b y=0$
$$
\begin{aligned}
& \Rightarrow \quad a x+b y=0 \\
& P A=\frac{a^2}{\sqrt{a^2+b^2}} \\
& \therefore \quad m=\left|\frac{a^2}{\sqrt{a^2+b^2}}\right| \\
& \text { and } B Q=n=\left|\frac{b^2}{\sqrt{a^2+b^2}}\right| \\
& m^2+n^2=\frac{a^4+b^4}{a^2+b^2} \\
& m n=\frac{a^2 b^2}{a^2+b^2} \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \because \quad m^2+n^2+2 m n=\frac{a^4+b^4+2 a^2 b^2}{a^2+b^2} \\
& \qquad \begin{aligned}
(m+n)^2=\frac{\left(a^2+b^2\right)^2}{a^2+b^2}=a^2+b^2
\end{aligned} \\
& \therefore \text { Diametre of circle }=\sqrt{a^2+b^2}=\sqrt{(m+n)^2} \\
& =m+n
\end{aligned}
$$
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