JEE MAIN - Mathematics (2024 - 4th April Evening Shift - No. 14)
Let $$\mathrm{C}$$ 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
$$3-\sqrt{2}$$
$$2-\sqrt{3}$$
$$\sqrt{2}-1$$
$$\sqrt{2}+1$$
Explanation
$$\text { Let the line by } x+y=\lambda$$
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$$\begin{aligned} & \therefore\left|\frac{\lambda}{\sqrt{2}}\right|=3 \\ & \therefore \quad \lambda= \pm 3 \sqrt{2} \end{aligned}$$
$$\therefore$$ distance between lines $$x+y=2$$ and $$x+y=3 \sqrt{2}$$ is
$$\frac{3 \sqrt{2}-2}{\sqrt{2}}=3-\sqrt{2}$$
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