Circle Ques 38
- 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: 38.(c)
Solution:
Formula:
- 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} $
BETA
