Question

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:

• A. $x^2+y^2-2xy=0$
• B. $x^2+y^2-2x^2{y}^2=0$
• C. $x^2+y^2-4x^2{y}^2=0$
• D. $x^2+y^2-16x^2{y}^2=0$

Answer 1

The correct option is C $x^2+y^2-4x^2{y}^2=0$

Let mid-point of $PQ$ be $S$ having coordinates $(h,k)$.
Hence, coordinates of $P$ and $Q$ are $(2h,0)\text{and}(0,2k)$ respectively.
Also, the equation of $PQ$ using Intercept form is
$\frac{x}{2h}+\frac{y}{2k}=1$
Given circle is $x^2+y^2=1$
$C=(0,0),r=1$
Also, $PQ$ is tangent to circle, so distance from centre to $PQ$ is equal to radius
$\left|\frac{-1}{\sqrt{{\left(\frac{1}{2h}\right)}^2+{\left(\frac{1}{2k}\right)}^2}}\right|=1\Rightarrow \frac{1}{4h^2}+\frac{1}{4k^2}=1\Rightarrow k^2+h^2=4h^2{k}^2$

Hence, the locus is
$x^2+y^2-4x^2{y}^2=0$