If a tangent to the circle
Question:

If a tangent to the circle $x^{2}+y^{2}=$ lintersects the coordinate axes at distinct points $P$ and $Q$, then the locus of the midpoint of $\mathrm{PQ}$ is:

1. (1) $x^{2}+y^{2}-4 x^{2} y^{2}=0$

2. (2) $x^{2}+y^{2}-2 x y=0$

3. (3) $x^{2}+y^{2}-16 x^{2} y^{2}=0$

4. (4) $x^{2}+y^{2}-2 x^{2} y^{2}=0$

Correct Option: 1,

Solution:

Let any tangent to circle $x^{2}+y^{2}=1$ is $x \cos \theta+y \sin \theta=1$

Since, $P$ and $Q$ are the point of intersection on the coordinate axes.

Then $P \equiv\left(\frac{1}{\cos \theta}, 0\right) \& Q \equiv\left(0, \frac{1}{\sin \theta}\right)$

$\therefore$ mid-point of PQ be $M \equiv\left(\frac{1}{2 \cos \theta}, \frac{1}{2 \sin \theta}\right) \equiv(h, k)$

$\Rightarrow \cos \theta=\frac{1}{2 h}$……….(1)

$\sin \theta=\frac{1}{2 k}$ ………….(2)

Now squaring and adding equation (1) and (2)

$\frac{1}{h^{2}}+\frac{1}{k^{2}}=4$

$\Rightarrow \mathrm{h}^{2}+\mathrm{k}^{2}=4 \mathrm{~h}^{2} \mathrm{k}^{2}$

$\therefore$ locus of $M$ is $: x^{2}+y^{2}=4 x^{2} y^{2}$