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

Answer 1

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

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)
$\therefore$ OR$=1\Rightarrow \left|\frac{-1}{\sqrt{{\left(\frac{1}{2h}\right)}^2+{\left(\frac{1}{2k}\right)}^2}}\right|=1$
$\Rightarrow \frac{1}{4h^2}+\frac{1}{4k^2}=1$
$\Rightarrow x^2+y^2-4x^2{y}^2=0$
Aliter:
tangent to circle
$x\cos \theta +y\sin \theta =1$
$P:(\sec \theta ,0)$
$Q:(0,cosec\theta )$
$2h=\sec \theta \Rightarrow \cos \theta =\frac{1}{2h}$ & $\sin \theta =\frac{1}{2k}$
$\frac{1}{(2x{)}^2}+\frac{1}{(2y{)}^2}=1$.