Key Concepts and Formulas

• Equation of a circle: A circle centered at the origin with radius $r$ has the equation $x^2 + y^2 = r^2$.
• Equation of a tangent to a circle: The equation of the tangent to the circle $x^2 + y^2 = r^2$ at the point $(x_1, y_1)$ on the circle is given by $xx_1 + yy_1 = r^2$.
• Midpoint Formula: The midpoint $M(h, k)$ of a line segment joining two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ is given by $h = \frac{x_1 + x_2}{2}$ and $k = \frac{y_1 + y_2}{2}$.
• Trigonometric Identity: The fundamental identity $\sin^2\theta + \cos^2\theta = 1$ is crucial for eliminating the parameter.

Step-by-Step Solution

Step 1: Parametrize a point on the circle. Let $(\cos\theta, \sin\theta)$ be a point on the circle $x^2 + y^2 = 1$.

Why this step? Using the parametric form allows us to represent any point on the circle using a single variable $\theta$, which simplifies finding the tangent equation and later eliminating it.

Step 2: Find the equation of the tangent at the point $(\cos\theta, \sin\theta)$. The equation of the tangent to the circle $x^2 + y^2 = 1$ at the point $(\cos\theta, \sin\theta)$ is given by $x\cos\theta + y\sin\theta = 1$.

Why this step? The tangent line is the central geometric element connecting the circle to the points P and Q, and subsequently to the midpoint M.

Step 3: Find the coordinates of the points P and Q where the tangent intersects the coordinate axes.

• Point P (on the x-axis): For a point on the x-axis, the y-coordinate is 0. Substitute $y=0$ into the tangent equation: $x\cos\theta + 0\cdot\sin\theta = 1$ $x\cos\theta = 1$ Since the problem states that P and Q are distinct points on the axes, the tangent cannot be parallel to an axis. Thus, $\cos\theta \neq 0$, so we can write: $x = \frac{1}{\cos\theta} = \sec\theta$ Thus, $P(\sec\theta, 0)$.

• Point Q (on the y-axis): For a point on the y-axis, the x-coordinate is 0. Substitute $x=0$ into the tangent equation: $0\cdot\cos\theta + y\sin\theta = 1$ $y\sin\theta = 1$ Since the problem states that P and Q are distinct points on the axes, the tangent cannot be parallel to an axis. Thus, $\sin\theta \neq 0$, so we can write: $y = \frac{1}{\sin\theta} = \csc\theta$ Thus, $Q(0, \csc\theta)$.

Why this step? P and Q define the segment whose midpoint we are interested in. Finding their coordinates in terms of $\theta$ is essential for using the midpoint formula.

Step 4: Express the coordinates of the midpoint M(h, k) of PQ. Let $M(h, k)$ be the midpoint of the segment PQ. Using the midpoint formula with $P(\sec\theta, 0)$ and $Q(0, \csc\theta)$: $h = \frac{\sec\theta + 0}{2} \quad \Rightarrow \quad h = \frac{\sec\theta}{2}$ $k = \frac{0 + \csc\theta}{2} \quad \Rightarrow \quad k = \frac{\csc\theta}{2}$

Why this step? This step establishes the relationship between the coordinates of the locus point $(h, k)$ and the parameter $\theta$. Our goal is to find an equation relating $h$ and $k$ without $\theta$.

Step 5: Eliminate the parameter $\theta$ using a trigonometric identity. From the expressions for $h$ and $k$, we can isolate $\sec\theta$ and $\csc\theta$: $\sec\theta = 2h \quad \Rightarrow \quad \cos\theta = \frac{1}{2h}$ $\csc\theta = 2k \quad \Rightarrow \quad \sin\theta = \frac{1}{2k}$ Now, we use the fundamental trigonometric identity $\sin^2\theta + \cos^2\theta = 1$. Substitute the expressions for $\sin\theta$ and $\cos\theta$: $\left(\frac{1}{2k}\right)^2 + \left(\frac{1}{2h}\right)^2 = 1$

Why this step? This is the crucial step to find the locus. We have two equations involving $h, k, \sin\theta, \cos\theta$. The identity $\sin^2\theta + \cos^2\theta = 1$ allows us to combine them and eliminate $\theta$.

Step 6: Simplify the equation to get the locus. $\frac{1}{4k^2} + \frac{1}{4h^2} = 1$ Multiply the entire equation by $4h^2k^2$: $h^2 + k^2 = 4h^2k^2$ Finally, to express the locus, replace $h$ with $x$ and $k$ with $y$: $x^2 + y^2 = 4x^2y^2$ Rearranging to match the options: $x^2 + y^2 - 4x^2y^2 = 0$

This matches option (A).

Common Mistakes & Tips

• Parametric Form: Using the parametric form simplifies the calculations significantly compared to using $(x_1, y_1)$ and the constraint $x_1^2 + y_1^2 = 1$.
• Conditions: Pay attention to given conditions. Here, "distinct points" implies $\cos\theta \neq 0$ and $\sin\theta \neq 0$.
• Algebra: Be careful with algebraic manipulations.

Summary

This problem demonstrates how to find the locus of a point using a parameter. The general approach is to express the coordinates of the point in terms of a parameter, eliminate the parameter using a known identity, and simplify the resulting equation. The final locus is $x^2 + y^2 - 4x^2y^2 = 0$.

The final answer is $\boxed{x^2 + y^2 - 4x^2y^2 = 0}$, which corresponds to option (A).