Mathematics - Circle Question with Solution | TestHub

Any point on the circle $x^2+y^2=r^2$ is $\left(r\cos \theta , r\sin \theta \right)$ and the equation of the tangent at this point is $x\cos \theta +y\sin \theta =r.$

Thus, any point on circle $x^2+y^2=1$ is $\left(\cos \theta , \sin \theta \right)$ and the equation of tangent at this point is $x\cos \theta +y\sin \theta =1.$

To find the point where this line intersects the $x$-axis, put $y=0,$$\Rightarrow x=\frac{1}{\cos \theta }$

$\Rightarrow P=\left(\frac{1}{\cos \theta }, 0\right)$

And, to find the point where this line intersects the $y$-axis, put $x=0,$ 

$\Rightarrow y=\frac{1}{\sin \theta }$

$\Rightarrow Q=\left(0, \frac{1}{\sin \theta }\right)$

The mid-point of a line segment joining the points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

Let, midpoint of $PQ$ is $\left(h, k\right)$

$\therefore \left(h, k\right)=\left(\frac{1}{2\cos \theta }, \frac{1}{2\sin \theta }\right)$

$\Rightarrow h=\frac{1}{2\cos \theta } \& k=\frac{1}{2\sin \theta }$

$\Rightarrow \cos \theta =\frac{1}{2h} \& \sin \theta =\frac{1}{2k}$

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

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

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

To get the locus of the required point, replace $\left(h, k\right)$ by $\left(x, y\right)$

$\therefore$Locus is $\frac{1}{x^2}+\frac{1}{y^2}=4$

$\Rightarrow x^2+y^2–4x^2{y}^2=0.$