Circle 2 Question 2

2. 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 $P Q$ is

(2019 Main, 9 April I)

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

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

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

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

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Answer:

Correct Answer: 2. (c)

Solution:

  1. Equation of given circle is $x^{2}+y^{2}=1$, then equation of tangent at the point $(\cos \theta, \sin \theta)$ on the given circle is

$ x \cos \theta+y \sin \theta=1 $

$[\because$ Equation of tangent at the point $P(\cos \theta, \sin \theta)$ to the circle $x^{2}+y^{2}=r^{2}$ is $\left.x \cos \theta+y \sin \theta=r\right]$

Now, the point of intersection with coordinate axes are $P(\sec \theta, 0)$ and $Q(0, \operatorname{cosec} \theta)$.

$\because$ Mid-point of line joining points $P$ and $Q$ is

$ M \frac{\sec \theta}{2}, \frac{\operatorname{cosec} \theta}{2}=(h, k) \text { (let) } $

So, $\cos \theta=\frac{1}{2 h}$ and $\sin \theta=\frac{1}{2 k}$

$\because \quad \sin ^{2} \theta+\cos ^{2} \theta=1$

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

Now, locus of mid-point $M$ is

$ \begin{aligned} \frac{1}{x^{2}}+\frac{1}{y^{2}} & =4 \\ \Rightarrow \quad x^{2}+y^{2}-4 x^{2} y^{2} & =0 \end{aligned} $



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