The locus of the midpoint of a chord of the c | vedclass

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(C) Let the midpoint be $(h, k)$. The equation of the chord with midpoint $(h, k)$ is $hx + ky = h^2 + k^2$.Homogenizing the circle equation $x^2 + y^

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$x^2 + y^2 = 2$

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(C) Let the midpoint be $(h, k)$. The equation of the chord with midpoint $(h, k)$ is $hx + ky = h^2 + k^2$.Homogenizing the circle equation $x^2 + y^2 = 4$ using the chord equation:$x^2 + y^2 = 4 \left( \frac{hx + ky}{h^2 + k^2} \right)^2$.Since the chord subtends a right angle at the origin,the sum of the coefficients of $x^2$ and $y^2$ must be zero.$1 - \frac{4h^2}{(h^2 + k^2)^2} + 1 - \frac{4k^2}{(h^2 + k^2)^2} = 0$.$2 - \frac{4(h^2 + k^2)}{(h^2 + k^2)^2} = 0$.$2 - \frac{4}{h^2 + k^2} = 0$.$h^2 + k^2 = 2$.Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 = 2$.

The locus of the midpoint of a chord of the circle $x^2 + y^2 = 4$ which subtends a right angle at the origin is

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