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
B
7/2
C
1
D
2

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 $$\dfrac{7}{2}.$$Maths

Suggest Corrections
Â
0

Similar questions
View More

People also searched for
View More