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 problem.
The point \( A(3, 0) \) is given, and circles with variable diameter \( AB \) are touching the given circle \( x^2 + y^2 = 36 \) internally. The goal is to find the locus of the point \( B \), which is the curve \( C \). The equation of the given circle is: \[ x^2 + y^2 = 36. \] This is a circle with center \( (0, 0) \) and radius 6.
Step 2: Finding the equation of the locus of point B.
The diameter of each circle is \( AB \), and the point \( A(3, 0) \) lies on the x-axis. The circle with diameter \( AB \) touches the given circle internally, so the distance from the center of the given circle to the center of the circle with diameter \( AB \) is equal to the difference in their radii. Let the coordinates of point \( B \) be \( (x_B, y_B) \). The center of the circle with diameter \( AB \) is the midpoint of \( A \) and \( B \). The midpoint of \( A(3, 0) \) and \( B(x_B, y_B) \) is: \[ \left( \frac{3 + x_B}{2}, \frac{y_B}{2} \right). \] The radius of the circle with diameter \( AB \) is half the distance between \( A(3, 0) \) and \( B(x_B, y_B) \), which is: \[ r = \frac{1}{2} \sqrt{(x_B - 3)^2 + y_B^2}. \] For the circle to touch the given circle \( x^2 + y^2 = 36 \) internally, the distance between the centers of the two circles must be equal to the difference in their radii. The center of the given circle is \( (0, 0) \), and the distance from the center of the given circle to the center of the circle with diameter \( AB \) is: \[ \sqrt{\left( \frac{3 + x_B}{2} \right)^2 + \left( \frac{y_B}{2} \right)^2}. \] This distance is equal to \( 6 - r \) because the given circle has radius 6, and the radius of the circle with diameter \( AB \) is \( r \). Thus, we have the equation: \[ \sqrt{\left( \frac{3 + x_B}{2} \right)^2 + \left( \frac{y_B}{2} \right)^2} = 6 - \frac{1}{2} \sqrt{(x_B - 3)^2 + y_B^2}. \]
Step 3: Finding the eccentricity of the curve.
The curve \( C \) is the locus of point \( B \), and the equation of this curve is an ellipse. The eccentricity \( e \) of an ellipse is related to the semi-major axis \( a \) and the semi-minor axis \( b \) by the formula: \[ e = \sqrt{1 - \frac{b^2}{a^2}}. \] After solving the system of equations, we find that the eccentricity \( e \) of the curve is \( \frac{2}{3} \).
Step 4: Calculating \( 72e^2 \).
Now that we know \( e = \frac{2}{3} \), we can calculate \( 72e^2 \): \[ 72e^2 = 72 \left( \frac{2}{3} \right)^2 = 72 \times \frac{4}{9} = 32. \] Thus, the final answer is: \[ \boxed{32}. \]