If a tangent to the circle x 2 + y 2 = 1 intersects the coordinate axes at distinct points P and Q , then…
If a tangent to the circle intersects the coordinate axes at distinct points and then the locus of the mid-point of is:
Solution
Any point on the circle $x^{2} + y^{2} = r^{2}$ is $(r \cos \theta, r \sin \theta)$ 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 $(\cos \theta, \sin \theta)$ 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 $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $\left(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}\right)$.
Let, midpoint of $PQ$ is $(h, k)$.
$\Rightarrow (h, k) = \left(\frac{1}{2 \cos \theta}, \frac{1}{2 \sin \theta}\right)$.
$\Rightarrow h = \frac{1}{2 \cos \theta}$ and $k = \frac{1}{2 \sin \theta}$.
$\Rightarrow \cos \theta = \frac{1}{2h}$ and $\sin \theta = \frac{1}{2k}$.
$\Rightarrow \sin^{2} \theta + \cos^{2} \theta = 1$.
$\Rightarrow \frac{1}{4h^{2}} + \frac{1}{4k^{2}} = 1$.
$\Rightarrow \frac{1}{h^{2}} + \frac{1}{k^{2}} = 4$.
To get the locus of the required point, replace $(h, k)$ by $(x, y)$.
$\Rightarrow$ Locus is $\frac{1}{x^{2}} + \frac{1}{y^{2}} = 4$.
$\Rightarrow x^{2} + y^{2} - 4x^{2}y^{2} = 0$.