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: Take coordinates of the variable point \( B \).
Let \( B(x,y) \).
The circle drawn on \( AB \) as diameter has center at the midpoint of \( A(3,0) \) and \( B(x,y) \).
So, its center is
\[ M\left(\frac{x+3}{2},\frac{y}{2}\right). \] Its radius is
\[ r=\frac{AB}{2}=\frac{1}{2}\sqrt{(x-3)^2+y^2}. \]
Step 2: Use the touching 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 must be equal to \( 6 \).
Thus,
\[ OM+r=6. \] Now,
\[ OM=\sqrt{\left(\frac{x+3}{2}\right)^2+\left(\frac{y}{2}\right)^2} =\frac{1}{2}\sqrt{(x+3)^2+y^2}. \] Hence,
\[ \frac{1}{2}\sqrt{(x+3)^2+y^2}+\frac{1}{2}\sqrt{(x-3)^2+y^2}=6. \] Multiplying by \( 2 \), we get
\[ \sqrt{(x+3)^2+y^2}+\sqrt{(x-3)^2+y^2}=12. \]
Step 3: Identify the locus.
The equation
\[ \sqrt{(x+3)^2+y^2}+\sqrt{(x-3)^2+y^2}=12 \] represents an ellipse.
Its foci are
\[ F_1(-3,0), \qquad F_2(3,0). \] So,
\[ 2a=12 \Rightarrow a=6. \] Also, the distance of each focus from the center is
\[ c=3. \]
Step 4: Find the eccentricity.
For an ellipse, eccentricity is
\[ e=\frac{c}{a}. \] Therefore,
\[ e=\frac{3}{6}=\frac{1}{2}. \] Final Answer: \( \dfrac{1}{2} \)