The locus of the midpoints of the chords of t | vedclass

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(B) Let the midpoint of the chord $AB$ be $C(x_{1}, y_{1})$. The origin is $O(0, 0)$.Since the chord $AB$ subtends a right angle at the origin,$\angle

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$x^{2}+y^{2}=2$

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(B) Let the midpoint of the chord $AB$ be $C(x_{1}, y_{1})$. The origin is $O(0, 0)$.Since the chord $AB$ subtends a right angle at the origin,$\angle AOB = 90^{\circ}$.In $\Delta OAB$,$OA = OB = r = 2$ (radius of the circle).Since $C$ is the midpoint of $AB$,$OC \perp AB$.In $\Delta OCB$,$\angle COB = \frac{1}{2} \angle AOB = 45^{\circ}$.Using trigonometry in $\Delta OCB$:$\cos(45^{\circ}) = \frac{OC}{OB}$$\frac{1}{\sqrt{2}} = \frac{\sqrt{x_{1}^{2} + y_{1}^{2}}}{2}$Squaring both sides:$\frac{1}{2} = \frac{x_{1}^{2} + y_{1}^{2}}{4}$$x_{1}^{2} + y_{1}^{2} = 2$Replacing $(x_{1}, y_{1})$ with $(x, y)$,the locus is $x^{2} + y^{2} = 2$.

The locus of the midpoints of the chords of the circle $x^{2}+y^{2}=4$ which subtend a right angle at the origin is

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