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Question

let $$P$$ and $$Q$$ are any two points on the circle $$x^{2}+y^{2}=4$$ such that $$PQ$$ is a diameter. If $$L_{1}$$ and $$L_{2}$$ are the lengths of perpendicular from $$P$$ and $$Q$$ on $$x+y=1$$, then the maximum value of $$L_{1},L_{2}$$ is


A
1/2
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B
7/2
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C
1
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D
2
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Solution

The correct option is D $$7/2$$
Let one end point of the circle be $$\left( 2\cos t, 2\sin t \right) $$ so the diametric point to it would be $$\left( 2\cos \left( 180+t\right), 2\sin \left( 180+t\right) \right)$$ that is $$\left( -2\cos t , -2\sin t \right).$$ The value of $$L_1\times L_2$$ would be -

$$\Rightarrow \left[ \dfrac{\left( 2\left( \cos t+\sin t\right) -1\right) }{\sqrt{2}}\right]\left[ \dfrac{2\left( \cos t+\sin t\right) +1 }{\sqrt{2}}\right]$$
$$\Rightarrow \left[ \dfrac{4\left( \cos t+\sin t\right)^2 -1 }{2}\right]$$
now, maximum value $$\left( \cos t+\sin t\right) $$ is $$\sqrt{2}$$
maximum value of $$L_1\times L_2=\dfrac{\left[ 4\left( 2\right) -1\right]}{2}$$

                                                $$=\dfrac{8-1}{2}$$
                                                $$=3.5$$ or $$\dfrac{7}{2}$$
Hence, the answer is $$\dfrac{7}{2}.$$





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