If a tangent to the circle x ^ 2 + y ^ 2 = 1 intersects the...

Any point on the circle ${x}^{2} + {y}^{2} = {r}^{2}$ is \left(rcos\theta, rsin\theta\right)$and the equation of the tangent at this point is$xcos\theta + ysin\theta = r .$</p><p>Thus, any point on circle${x}^{2} + {y}^{2} = 1$is \left(cos\theta, sin\theta\right)$ and the equation of tangent at this point is $xcos\theta + ysin\theta = 1 .$

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

$⇒ P = \left(\dfrac{1}{cos\theta}, 0\right)$

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

$⇒ y = \dfrac{1}{sin\theta}$

$⇒ Q = \left(0, \dfrac{1}{sin\theta}\right)$

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(\dfrac{{x}_{1} + {x}_{2}}{2}, \dfrac{{y}_{1} + {y}_{2}}{2}\right)$</p><p>Let, midpoint of$PQ$is \left(h, k\right)$

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

$⇒ h = \dfrac{1}{2cos\theta} \text{ and } k = \dfrac{1}{2sin\theta}$

$⇒ cos\theta = \dfrac{1}{2h} \text{ and } sin\theta = \dfrac{1}{2k}$

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

$⇒ \dfrac{1}{4{h}^{2}} + \dfrac{1}{4{k}^{2}} = 1$

$⇒ \dfrac{1}{{h}^{2}} + \dfrac{1}{{k}^{2}} = 4$

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

$∴$Locus is $\dfrac{1}{{x}^{2}} + \dfrac{1}{{y}^{2}} = 4$

$⇒ {x}^{2} + {y}^{2} – 4{x}^{2}{y}^{2} = 0 .$