Let $C$ be a circle with radius $\sqrt{10}$ units and centre at the origin. Let the line $x+y=2$ intersects…

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
  1. $\begin{aligned}& 3-\sqrt{2} \end{aligned}$
  2. $\begin{aligned} &\sqrt{2}+1 \end{aligned}$
  3. $\begin{aligned} &\sqrt{2}-1 \end{aligned}$
  4. $\begin{aligned}& 2-\sqrt{3} \end{aligned}$

Solution


$\begin{aligned} & C: x^2+y^2=10 \\ & A N=\frac{M N}{2}=1 \\ & \therefore \text { In } \triangle O A N \rightarrow(O N)^2=(O A)^2+(A N)^2 \\ & 10=(O A)^2+1 \rightarrow O A=3 \end{aligned}$
Perpendicular distance of center from $\mathrm{PQ}=\frac{|0+0-2|}{\sqrt{2}}=\sqrt{2}$
Perpendicular distance between $\mathrm{MN}$ and $\begin{aligned} & \mathrm{PQ}=\mathrm{OA}+\sqrt{2} \text { or }|\mathrm{OA}-\sqrt{2}| \\ & =3+\sqrt{2} \text { or } 3-\sqrt{2} \end{aligned}$

Asked in: JEE Main 2024 (04 Apr Shift 2)

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