Question

Let A be the point \( (3, 0) \) and circles with variable diameter AB touch the circle \( x^2 + y^2 = 36 \) internally. Let the curve \( C \) be the locus of the point B. If the eccentricity of \( C \) is \( e \), then \( 72e^2 \) is equal to _______.

Answer 1

Correct Answer: 32

Step 1: Understanding the Concept:
Let the given circle be \(S_1: x^2 + y^2 = 36\). Its center is \(O(0,0)\) and radius is \(R = 6\).
Let the variable circle be \(S_2\). Its diameter is \(AB\), so its center \(M\) is the midpoint of \(AB\) and its radius is \(r = MA\).
Since \(S_2\) touches \(S_1\) internally, the distance between their centers must be equal to the difference of their radii.
Step 2: Key Formula or Approach:
Condition for internal touching: \(OM = R - r\).
Here, \(R = 6\) and \(r = MA\), so \(OM = 6 - MA \implies OM + MA = 6\).
This means the sum of distances from \(M\) to two fixed points \(O\) and \(A\) is constant, which is the definition of an ellipse.
Step 3: Detailed Explanation:
The locus of \(M\) is an ellipse with foci at \(O(0,0)\) and \(A(3,0)\).
The length of the major axis of this ellipse is \(2a_M = 6 \implies a_M = 3\).
The distance between the foci is \(2c_M = OA = 3\).
The eccentricity of the locus of \(M\) is:
\[ e_M = \frac{2c_M}{2a_M} = \frac{3}{6} = \frac{1}{2} \] We are asked to find the eccentricity of the locus of \(B\).
Since \(M\) is the midpoint of \(AB\), we can write \(\vec{M} = \frac{\vec{A} + \vec{B}}{2} \implies \vec{B} = 2\vec{M} - \vec{A}\).
This equation represents a homothety (scaling and translation).
Transformations like scaling and translation do not change the shape or eccentricity of a conic section.
Therefore, the locus of \(B\) is an ellipse with the exact same eccentricity as the locus of \(M\).
So, \(e = \frac{1}{2}\).
We need to calculate the value of \(72e^2\).
\[ 72e^2 = 72 \left(\frac{1}{2}\right)^2 = 72 \left(\frac{1}{4}\right) = 18 \] Step 4: Final Answer:
The value of \(72e^2\) is \(18\).