Question

Let `C' be a circle with radius `6' units centred at origin. Let \( A(3,0) \) be a point. If \( B \) is a variable point in xy-plane such that circle drawn taking \( AB \) as diameter touches the circle \( C \), then eccentricity of the locus of point `B' is

• A. 2
• B. \( \dfrac{1}{2} \)
• C. 3
• D. \( \dfrac{3}{4} \)

Hint:

Whenever the sum of distances of a moving point from two fixed points is constant, the locus is an ellipse. Here the touching condition converts directly into that standard form.

Answer 1

The Correct Option is B

Step 1: Assume coordinates of the variable point \( B \).
Let the coordinates of point \( B \) be \( B(x,y) \). A circle is drawn with \( AB \) as its diameter, where \( A(3,0) \). The center of this circle is the midpoint of \( A \) and \( B \), given by \[ M\left(\frac{x+3}{2}, \frac{y}{2}\right). \] The radius of the circle is half the length of \( AB \): \[ r = \frac{AB}{2} = \frac{1}{2} \sqrt{(x-3)^2 + y^2}. \]
Step 2: Apply the internal tangency condition.
The given circle \( C \) has center \( O(0,0) \) and radius \( 6 \). Since the circle with diameter \( AB \) touches circle \( C \) internally, the distance between their centers plus the radius of the smaller circle equals the radius of the larger circle: \[ OM + r = 6. \] Now, compute the distance between the centers: \[ OM = \sqrt{ \left(\frac{x+3}{2}\right)^2 + \left(\frac{y}{2}\right)^2 } = \frac{1}{2} \sqrt{(x+3)^2 + y^2}. \] Substituting into the tangency condition: \[ \frac{1}{2} \sqrt{(x+3)^2 + y^2} + \frac{1}{2} \sqrt{(x-3)^2 + y^2} = 6. \] Multiplying both sides by \( 2 \): \[ \sqrt{(x+3)^2 + y^2} + \sqrt{(x-3)^2 + y^2} = 12. \]
Step 3: Recognize the geometric locus.
The equation \[ \sqrt{(x+3)^2 + y^2} + \sqrt{(x-3)^2 + y^2} = 12 \] represents an ellipse, where the sum of distances from two fixed points (foci) is constant. Thus, the foci are: \[ F_1(-3,0), \qquad F_2(3,0). \] From the standard form of an ellipse: \[ 2a = 12 \quad \Rightarrow \quad a = 6. \] The distance from the center to each focus is: \[ c = 3. \]
Step 4: Calculate the eccentricity.
For an ellipse, the eccentricity is defined as: \[ e = \frac{c}{a}. \] Substituting the values: \[ e = \frac{3}{6} = \frac{1}{2}. \] Conclusion:
Hence, the locus of point \( B \) forms an ellipse whose eccentricity is \( \dfrac{1}{2} \).
Final Answer: \( \dfrac{1}{2} \).