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Let the line $2 \mathrm{x}+3 \mathrm{y}-\mathrm{k}=0, \mathrm{k}>0$, intersect the $\mathrm{x}$-axis and $\mathrm{y}$-axis at the points $\mathrm{A}$ and $\mathrm{B}$, respectively. If the equation of the circle having the line segment $\mathrm{AB}$ as a diameter is $\mathrm{x}^2+\mathrm{y}^2-3 \mathrm{x}-2 \mathrm{y}=0$ and the length of the latus rectum of the ellipse $\mathrm{x}^2+9 \mathrm{y}^2=\mathrm{k}^2$ is $\frac{\mathrm{m}}{\mathrm{n}}$, where $\mathrm{m}$ and $\mathrm{n}$ are coprime, then $2 \mathrm{~m}+\mathrm{n}$ is equal to
$10$
$11$
$13$
$12$
Solution
Centre of the circle $=\left(\frac{3}{2}, 1\right) $
Equation of diameter $=2 \mathrm{x}+3 \mathrm{y}-\mathrm{k}=0 $
$ 2\left(\frac{3}{2}\right)+3(1)-\mathrm{k}=0 $
$ \Rightarrow \mathrm{k}=6$
Now, Equation of ellipse becomes
$ x^2+9 y^2=36 $
$ \frac{x^2}{6^2}+\frac{y^2}{2^2}=1$
length of $L R=\frac{2 b^2}{a}=\frac{2 \cdot 2^2}{6}=\frac{8}{6}=\frac{4}{3}=\frac{m}{n} $
$ \therefore 2 m+n=2(4)+3=11$
