JEE MAIN - Mathematics (2006 - No. 28)
Let $$C$$ be the circle with centre $$(0, 0)$$ and radius $$3$$ units. The equation of the locus of the mid points of the chords of the circle $$C$$ that subtend an angle of $${{2\pi } \over 3}$$ at its center is :
$${x^2} + {y^2} = {3 \over 2}$$
$${x^2} + {y^2} = 1$$
$${x^2} + {y^2} = {{27} \over 4}$$
$${x^2} + {y^2} = {{9} \over 4}$$
Explanation
Let $$M\left( {h,k} \right)$$ be the mid point of chord $$AB$$ where
$$\angle AOB = {{2\pi } \over 3}$$
$$\therefore$$ $$\angle AOM = {\pi \over 3}.$$ Also $$OM=$$ $$3\cos {\pi \over 3} = {3 \over 2}$$
$$ \Rightarrow \sqrt {{h^2} + k{}^2} = {3 \over 2}$$
$$ \Rightarrow {h^2} + {k^2} = {9 \over 4}$$
$$\therefore$$ Locus of $$\left( {h,k} \right)$$ is
$${x^2} + {y^2} = {9 \over 4}$$
$$\angle AOB = {{2\pi } \over 3}$$
$$\therefore$$ $$\angle AOM = {\pi \over 3}.$$ Also $$OM=$$ $$3\cos {\pi \over 3} = {3 \over 2}$$
$$ \Rightarrow \sqrt {{h^2} + k{}^2} = {3 \over 2}$$
$$ \Rightarrow {h^2} + {k^2} = {9 \over 4}$$
$$\therefore$$ Locus of $$\left( {h,k} \right)$$ is
$${x^2} + {y^2} = {9 \over 4}$$
Comments (0)
