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Question: Answered & Verified by Expert
The locus of the mid-points of chords of the circle $x^{2}+y^{2}=1$, which subtends a right angle at the origin, is
MathematicsCircleWBJEEWBJEE 2016
Options:
  • A $x^{2}+y^{2}=\frac{1}{4}$
  • B $x^{2}+y^{2}=\frac{1}{2}$
  • C $x y=0$
  • D $x^{2}-y^{2}=0$
Solution:
2538 Upvotes Verified Answer
The correct answer is: $x^{2}+y^{2}=\frac{1}{2}$
Let $(h, k)$ be the coordinates of the mid-point of a chord which subtends a right angle at the origin. Then, equation of the chord is
$$
\begin{array}{l}
\quad h x+h y-1=h^{2}+k^{2}-1 \quad[u \operatorname{sing} T=S] \\
\Rightarrow \quad h x+k y=h^{2}+k^{2}
\end{array}
$$
The combined equation of the pair of lines joining the origin to the points of intersection of $x^{2}+y^{2}=1$ and $h x+k y=h^{2}+k^{2}$ is
$$
x^{2}+y^{2}-1\left(\frac{h x+h y}{h^{2}+k^{2}}\right)^{2}=0
$$
Lines given by the above equation are at right angle. Therefore, coefficient of $x^{2}+$ coefficient of $y^{2}=0$
ie. $\quad h^{2}+k^{2}=\frac{1}{2}$
$\therefore x^{2}+y^{2}=\frac{1}{2}$

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