Understanding the Problem: The Interplay of Curves

This problem asks us to find the locus of points that satisfy two geometric conditions simultaneously. We are looking for the midpoints of chords of a given circle, $x^2 + y^2 = 25$. The crucial constraint is that these chords must also be tangent to a given hyperbola, $\frac{x^2}{9} - \frac{y^2}{16} = 1$. The core idea is to express the equation of this "special" line (the chord/tangent) in two different ways—once as a chord of the circle and once as a tangent to the hyperbola—and then equate their properties to find the relationship between the coordinates of the midpoint.

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Key Concepts and Formulas:

To solve this problem efficiently, we need to utilize specific formulas related to circles and hyperbolas:

1. Equation of a Chord of a Circle with a Given Midpoint: For a circle $x^2 + y^2 = r^2$, if $P(x_1, y_1)$ is the midpoint of a chord, the equation of that chord is given by the formula $T = S_1$.

   • $T$ represents the expression for the tangent at $(x_1, y_1)$ but used in the context of a chord: $xx_1 + yy_1 - r^2$.
   • $S_1$ represents the value of the circle's equation when $(x,y)$ is replaced by $(x_1, y_1)$: $x_1^2 + y_1^2 - r^2$.
   • Equating these gives: $xx_1 + yy_1 - r^2 = x_1^2 + y_1^2 - r^2$.
   • This simplifies to: $xx_1 + yy_1 = x_1^2 + y_1^2$ This formula is incredibly efficient for problems involving midpoints of chords, as it directly relates the coordinates of the midpoint to the equation of the chord.

2. Equation of a Tangent to a Hyperbola with a Given Slope: For the standard hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the equation of a tangent line with slope $m$ is given by: $y = mx \pm \sqrt{a^2m^2 - b^2}$ This formula allows us to express any tangent line to the hyperbola solely in terms of its slope $m$ and the hyperbola's parameters $a^2$ and $b^2$. The $\pm$ sign indicates that for a given non-vertical slope $m$, there are generally two such tangent lines (one on each "side" of the hyperbola, or one for each branch, depending on the slope).

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Step-by-Step Solution:

1. Define the Locus Point Let $P(h, k)$ be the midpoint of the chord of the circle $x^2 + y^2 = 25$. Our ultimate goal is to find an equation relating $h$ and $k$, which will define the locus of all such midpoints. We will replace $(h, k)$ with $(x, y)$ at the end.

2. Formulate the Equation of the Chord of the Circle The given circle is $x^2 + y^2 = 25$. Here, the radius squared is $r^2 = 25$. Using the $T = S_1$ formula for the chord with midpoint $(h, k)$: $xh + yk = h^2 + k^2 \quad \text{... (1)}$ Explanation: This equation represents the specific chord of the circle $x^2+y^2=25$ for which $(h,k)$ is the midpoint. Any point $(x,y)$ lying on this line belongs to the chord whose midpoint is $(h,k)$.

To facilitate comparison with the general tangent equation (which is in the slope-intercept form $y=mx+c$), we rearrange equation (1): $yk = -xh + (h^2 + k^2)$ $y = -\frac{h}{k}x + \frac{h^2 + k^2}{k} \quad \text{... (1')}$ Explanation: We isolate $y$ to clearly identify the slope $M = -\frac{h}{k}$ and the y-intercept $C = \frac{h^2+k^2}{k}$ of this chord. This rearrangement is crucial for the next step of equating coefficients. Note that this form assumes $k \neq 0$. We will address the $k=0$ case later to ensure our solution is comprehensive.

3. Formulate the Equation of the Tangent to the Hyperbola The given hyperbola is $\frac{x^2}{9} - \frac{y^2}{16} = 1$. Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we identify the parameters: $a^2 = 9$ $b^2 = 16$

Now, we write the equation of a tangent to this hyperbola with an arbitrary slope $m$: $y = mx \pm \sqrt{a^2m^2 - b^2}$ Substituting the values of $a^2$ and $b^2$: $y = mx \pm \sqrt{9m^2 - 16} \quad \text{... (2)}$ Explanation: This is the general form of any line that is tangent to the given hyperbola. The slope $m$ is a parameter that defines a specific tangent. The $\pm$ sign indicates that for a given slope $m$, there are two such tangent lines, one for each branch of the hyperbola (or sometimes two tangents for one branch).

4. Equate the Chord and Tangent Equations The problem statement implies that the chord of the circle is tangent to the hyperbola. This means that equation (1') and equation (2) represent the exact same line. If two lines are identical, their slopes must be equal, and their y-intercepts must also be equal.

Comparing the slopes: From (1'), the slope of the chord is $M = -\frac{h}{k}$. From (2), the slope of the tangent is $m$. Therefore: $m = -\frac{h}{k}$

Comparing the y-intercepts: From (1'), the y-intercept of the chord is $C = \frac{h^2 + k^2}{k}$. From (2), the y-intercept of the tangent is $\pm \sqrt{9m^2 - 16}$. Therefore: $\frac{h^2 + k^2}{k} = \pm \sqrt{9m^2 - 16}$ Explanation: This step is the core of the problem. By equating the corresponding properties (slope and y-intercept), we establish a relationship between the coordinates of the midpoint $(h,k)$ and the slope $m$ of the common line. This is how we link the conditions from the circle and the hyperbola.

5. Eliminate the Parameter 'm' and Simplify We now have two equations involving $h, k,$ and $m$. To find the locus of $(h, k)$, we must eliminate the parameter $m$. We substitute the expression for $m$ from the slope equality into the y-intercept equality: $\frac{h^2 + k^2}{k} = \pm \sqrt{9\left(-\frac{h}{k}\right)^2 - 16}$ $\frac{h^2 + k^2}{k} = \pm \sqrt{9\frac{h^2}{k^2} - 16}$ To eliminate the square root and the $\pm$ sign, we square both sides of the equation. Squaring handles both cases of the $\pm$ sign simultaneously, as $(\pm A)^2 = A^2$: $\left(\frac{h^2 + k^2}{k}\right)^2 = \left(\pm \sqrt{9\frac{h^2}{k^2} - 16}\right)^2$ $\frac{(h^2 + k^2)^2}{k^2} = 9\frac{h^2}{k^2} - 16$ Explanation: Squaring both sides is a standard algebraic technique to get rid of square roots and the ambiguity of the $\pm$ sign. We then combine the terms on the right-hand side using a common denominator to prepare for clearing the fraction.

Now, we combine the terms on the right-hand side: $\frac{(h^2 + k^2)^2}{k^2} = \frac{9h^2 - 16k^2}{k^2}$ Assuming $k \neq 0$, we can multiply both sides by $k^2$ to clear the denominators: $(h^2 + k^2)^2 = 9h^2 - 16k^2$ Explanation: Multiplying by $k^2$ simplifies the equation, leaving us with a clean relationship between $h$ and $k$. This equation defines the locus of the midpoint $(h,k)$.

6. Replace (h, k) with (x, y) to Find the Locus The equation $(h^2 + k^2)^2 = 9h^2 - 16k^2$ describes the relationship that any point $(h, k)$ satisfying the problem conditions must fulfill. To express this as the locus of a general point $(x, y)$, we simply replace $h$ with $x$ and $k$ with $y$: $(x^2 + y^2)^2 = 9x^2 - 16y^2$ Rearranging the terms to match the format of the given options: $(x^2 + y^2)^2 - 9x^2 + 16y^2 = 0$

This equation matches Option (A).

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Consideration of the $k=0$ Case (Special Case for Horizontal Chords/Vertical Tangents):

Our derivation in Step 2 and Step 5 assumed $k \neq 0$ when dividing by $k$. Let's check what happens if $k=0$. If the midpoint is $(h,0)$, the chord equation $xh + yk = h^2 + k^2$ becomes $xh = h^2$.

• If $h=0$, then $(h,k)=(0,0)$. The chord equation becomes $0=0$, which doesn't represent a unique line. The origin $(0,0)$ cannot be the midpoint of a chord that is also tangent to the hyperbola (as the hyperbola does not pass through the origin and its tangents don't pass through the origin's neighborhood in a way that would make it a midpoint).
• If $h \neq 0$, then $x=h$. This is a vertical line.

Can a vertical line be tangent to the hyperbola $\frac{x^2}{9} - \frac{y^2}{16} = 1$? Yes, vertical tangents to a hyperbola of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ exist at $x = \pm a$. For our hyperbola, $a^2 = 9$, so $a = \pm 3$. Thus, $x = \pm 3$ are the vertical tangents to the hyperbola. If $x=h$ is a vertical tangent, then $h = \pm 3$. So, the points $(\pm 3, 0)$ are potential midpoints. Let's substitute $(h,k) = (\pm 3, 0)$ into our derived locus equation: $((\pm 3)^2 + 0^2)^2 - 9(\pm 3)^2 + 16(0)^2 = 0$ $(9)^2 - 9(9) + 0 = 0$ $81 - 81 = 0$ $0 = 0$ This confirms that the points $(\pm 3, 0)$ (which correspond to vertical tangents $x=\pm 3$) are included in the locus. Therefore, our general derivation is valid, and the $k=0$ case is correctly incorporated by the final equation.

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Tips and Common Mistakes:

• Master the $T=S_1$ formula: This formula is a huge time-saver for chord midpoint problems. Make sure you understand its derivation and application.
• Don't forget the $\pm$ in the tangent equation: When equating y-intercepts, it's crucial to include the $\pm$ sign from the tangent formula. Squaring both sides is the correct way to handle it, as it removes both the square root and the sign.
• Parameter Elimination is Key: Locus problems often involve introducing a parameter (like slope $m$ or an angle $\theta$) that represents a variable property of the line or point. The core task is to systematically eliminate this parameter to get an equation purely in terms of the locus coordinates $(h, k)$.
• Careful Algebraic Manipulation: Squaring expressions, combining fractions, and clearing denominators require precision. A small error can lead to an incorrect locus. Double-check your algebra!
• Check for special cases (like $k=0$ or $m=0$): While the general derivation is often robust, quickly thinking about edge cases (like vertical or horizontal lines) can confirm your result or highlight potential exclusions from the locus. This adds confidence to your solution.

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Summary and Key Takeaway:

To find the locus of midpoints of chords of a circle that are tangent to a hyperbola, the strategy involves:

1. Define the Midpoint: Let the midpoint be $(h,k)$.
2. Chord Equation: Write the equation of the chord of the circle with $(h,k)$ as its midpoint using $T=S_1$, then convert it to slope-intercept form ($y=Mx+C$).
3. Tangent Equation: Write the general equation of a tangent to the hyperbola with slope $m$ ($y=mx \pm \sqrt{a^2m^2-b^2}$).
4. Equate and Eliminate: Since the chord is the tangent, equate their slopes and y-intercepts. Then, eliminate the parameter $m$ by substitution and algebraic manipulation (typically by squaring both sides).
5. Final Locus: Replace $(h,k)$ with $(x,y)$ to obtain the locus equation.

The resulting locus for this problem is a