Locus of Midpoint of Line Segment Cut Off by Tangent to Circle

To find the locus of the mid-point of the segment $MN$, we proceed step-by-step:

Step 1: Parametric representation of point $P$

The given equation of the circle $S$ is: $x^2 + y^2 = 4$

Since point $P$ lies on the circle $S$ and both of its coordinates are positive, we can express $P$ in parametric form as: $P = (2\cos\theta, 2\sin\theta), \quad \text{where } \theta \in \left(0, \frac{\pi}{2}\right)$

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Step 2: Equation of the tangent at $P$

The equation of the tangent to the circle $x^2 + y^2 = r^2$ at the point $(x_1, y_1)$ is given by $x x_1 + y y_1 = r^2$.

Substituting the coordinates of $P(2\cos\theta, 2\sin\theta)$ and $r^2 = 4$: $x(2\cos\theta) + y(2\sin\theta) = 4$ $x\cos\theta + y\sin\theta = 2$

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Step 3: Finding the coordinates of $M$ and $N$

The tangent intersects the coordinate axes at points $M$ and $N$:

• For the $x$-intercept (point $M$), set $y = 0$: $x\cos\theta = 2 \implies x = \frac{2}{\cos\theta}$ $\implies M = \left(\frac{2}{\cos\theta}, 0\right)$

• For the $y$-intercept (point $N$), set $x = 0$: $y\sin\theta = 2 \implies y = \frac{2}{\sin\theta}$ $\implies N = \left(0, \frac{2}{\sin\theta}\right)$

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Step 4: Mid-point of the segment $MN$

Let $(h, k)$ be the coordinates of the mid-point of the line segment $MN$. Using the midpoint formula: $h = \frac{\frac{2}{\cos\theta} + 0}{2} = \frac{1}{\cos\theta}$ $k = \frac{0 + \frac{2}{\sin\theta}}{2} = \frac{1}{\sin\theta}$

From these equations, we get: $\cos\theta = \frac{1}{h}$ $\sin\theta = \frac{1}{k}$

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Step 5: Eliminating the parameter $\theta$

Using the fundamental trigonometric identity $\sin^2\theta + \cos^2\theta = 1$: $\left(\frac{1}{k}\right)^2 + \left(\frac{1}{h}\right)^2 = 1$ $\frac{1}{h^2} + \frac{1}{k^2} = 1$

Taking the common denominator: $\frac{h^2 + k^2}{h^2 k^2} = 1 \implies h^2 + k^2 = h^2 k^2$

Replacing $(h, k)$ with $(x, y)$ to write the equation of the locus: $x^2 + y^2 = x^2 y^2$

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Conclusion

The mid-point of the segment $MN$ lies on the curve $x^2 + y^2 = x^2 y^2$.

Correct Option: (D)