• # question_answer If a tangent to the circle${{x}^{2}}+{{y}^{2}}=1$intersects the coordinate axes at distinct points P and Q, then the locus of the mid-point of PQ is [JEE Main 9-4-2019 Morning] A) ${{x}^{2}}+{{y}^{2}}2xy=0$B) ${{x}^{2}}+{{y}^{2}}16{{x}^{2}}{{y}^{2}}=0$C) ${{x}^{2}}+{{y}^{2}}4{{x}^{2}}{{y}^{2}}=0$D) ${{x}^{2}}+{{y}^{2}}2{{x}^{2}}{{y}^{2}}=0$

Let the mid point be S(h,k) $\therefore$P(2h,0) and Q(0,2k) equation of  $PQ:\frac{x}{2h}+\frac{y}{2k}=1$ $\because PQ$is tangent to circle at R(say) $\therefore$ $\Rightarrow \frac{1}{4{{h}^{2}}}+\frac{1}{4{{k}^{2}}}=1$ $\Rightarrow {{x}^{2}}+{{y}^{2}}-4{{x}^{2}}{{y}^{2}}=0$ Aliter : tangent to circle $x\cos \theta +y\sin \theta =1$ $P:(sec\theta ,0)$ $Q:(0,cosec\theta )$ $2h=\sec \theta \Rightarrow \cos \theta =\frac{1}{2h}\And \sin \theta =\frac{1}{2k}$ $\frac{1}{{{\left( 2x \right)}^{2}}}+\frac{1}{{{\left( 2y \right)}^{2}}}=1$