The locus of the mid-points of chords of the | vedclass

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(B) Let $(h, k)$ be the coordinates of the mid-point of a chord of the circle $x^{2}+y^{2}=1$. The equation of the chord with mid-point $(h, k)$ is gi

Vedclass · Accepted Answer

$x^{2}+y^{2}=\frac{1}{2}$

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(B) Let $(h, k)$ be the coordinates of the mid-point of a chord of the circle $x^{2}+y^{2}=1$. The equation of the chord with mid-point $(h, k)$ is given by $T=S_1$, which is $hx+ky = h^{2}+k^{2}$.The equation of the pair of lines joining the origin to the points of intersection of the circle and the chord is obtained by homogenizing the circle equation using the chord equation:$x^{2}+y^{2} = 1 \cdot \left(\frac{hx+ky}{h^{2}+k^{2}}ight)^{2}$$(h^{2}+k^{2})^{2}(x^{2}+y^{2}) = (hx+ky)^{2}$$(h^{2}+k^{2})^{2}(x^{2}+y^{2}) = h^{2}x^{2} + k^{2}y^{2} + 2hkxy$Since the chord subtends a right angle at the origin, the sum of the coefficients of $x^{2}$ and $y^{2}$ must be zero:$(h^{2}+k^{2})^{2} - h^{2} + (h^{2}+k^{2})^{2} - k^{2} = 0$$2(h^{2}+k^{2})^{2} - (h^{2}+k^{2}) = 0$Since $h^{2}+k^{2} 
eq 0$, we have $2(h^{2}+k^{2}) = 1$, or $h^{2}+k^{2} = \frac{1}{2}$.Replacing $(h, k)$ with $(x, y)$, the locus is $x^{2}+y^{2} = \frac{1}{2}$.

The locus of the mid-points of chords of the circle $x^{2}+y^{2}=1$, which subtend a right angle at the origin, is

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