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Let $C$ be a circle with radius $\sqrt{10}$ units and centre at the origin. Let the line $x+y=2$ intersects the circle $\mathrm{C}$ at the points $\mathrm{P}$ and $\mathrm{Q}$. Let $\mathrm{MN}$ be a chord of $\mathrm{C}$ of length 2 unit and slope -1. Then, a distance (in units) between the chord PQ and the chord $\mathrm{MN}$ is
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The correct answer is:
$\begin{aligned}& 3-\sqrt{2} \end{aligned}$
$\begin{aligned}
& C: x^2+y^2=10 \\
& A N=\frac{M N}{2}=1 \\
& \therefore \text { In } \triangle O A N \rightarrow(O N)^2=(O A)^2+(A N)^2 \\
& 10=(O A)^2+1 \rightarrow O A=3
\end{aligned}$
Perpendicular distance of center from
$\mathrm{PQ}=\frac{|0+0-2|}{\sqrt{2}}=\sqrt{2}$
Perpendicular distance between $\mathrm{MN}$ and
$\begin{aligned}
& \mathrm{PQ}=\mathrm{OA}+\sqrt{2} \text { or }|\mathrm{OA}-\sqrt{2}| \\
& =3+\sqrt{2} \text { or } 3-\sqrt{2}
\end{aligned}$
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