The area (in sq. units) of the locus of the point at which the two circles `x^(2)+y^(2)=1` and `(x-4)^(2)+y^(2)=4` subtend equal angles is

Options

A. `(32//9)pi`
B. `(32//3)pi`
C. `(64//3)pi`
D. `(64//9)pi`

Answer

To solve the problem, we need to find the area of the locus of points from which the two circles subtend equal angles. The circles given are:

1. Circle C1: \(x^2 + y^2 = 1\) (center at (0,0) and radius \(R_1 = 1\))
2. Circle C2: \((x - 4)^2 + y^2 = 4\) (center at (4,0) and radius \(R_2 = 2\))

### Step-by-step Solution:

#### Step 1: Understanding the condition of equal angles
The condition that the two circles subtend equal angles at a point \(P(h, k)\) means that the tangents drawn from point \(P\) to both circles will form equal angles with the line joining \(P\) to the centers of the circles.

#### Step 2: Finding the lengths of tangents
The length of the tangent from point \(P(h, k)\) to circle \(C_1\) is given by:
\[ s_1 = \sqrt{h^2 + k^2 - 1} \]
And for circle \(C_2\):
\[ s_2 = \sqrt{(h - 4)^2 + k^2 - 4} \]

#### Step 3: Setting up the tangent condition
Since the angles subtended are equal, we have:
\[ \frac{R_1}{s_1} = \frac{R_2}{s_2} \]
Substituting the values of \(R_1\) and \(R_2\):
\[ \frac{1}{\sqrt{h^2 + k^2 - 1}} = \frac{2}{\sqrt{(h - 4)^2 + k^2 - 4}} \]

#### Step 4: Cross-multiplying and squaring
Cross-multiplying gives:
\[ \sqrt{(h - 4)^2 + k^2 - 4} = 2\sqrt{h^2 + k^2 - 1} \]
Squaring both sides results in:
\[ (h - 4)^2 + k^2 - 4 = 4(h^2 + k^2 - 1) \]

#### Step 5: Expanding and simplifying
Expanding both sides:
\[ h^2 - 8h + 16 + k^2 - 4 = 4h^2 + 4k^2 - 4 \]
Combining like terms:
\[ h^2 + k^2 - 8h + 12 = 4h^2 + 4k^2 \]
Rearranging gives:
\[ 0 = 3h^2 + 3k^2 + 8h - 12 \]
Dividing through by 3:
\[ h^2 + k^2 + \frac{8}{3}h - 4 = 0 \]

#### Step 6: Completing the square
To complete the square for \(h\):
\[ h^2 + \frac{8}{3}h = \left(h + \frac{4}{3}\right)^2 - \frac{16}{9} \]
Substituting back:
\[ \left(h + \frac{4}{3}\right)^2 + k^2 - \frac{16}{9} - 4 = 0 \]
This simplifies to:
\[ \left(h + \frac{4}{3}\right)^2 + k^2 = \frac{16}{9} + \frac{36}{9} = \frac{52}{9} \]

#### Step 7: Identifying the locus
The equation \(\left(h + \frac{4}{3}\right)^2 + k^2 = \frac{52}{9}\) represents a circle centered at \(\left(-\frac{4}{3}, 0\right)\) with radius \(\sqrt{\frac{52}{9}} = \frac{2\sqrt{13}}{3}\).

#### Step 8: Finding the area
The area \(A\) of a circle is given by:
\[ A = \pi r^2 \]
Substituting the radius:
\[ A = \pi \left(\frac{2\sqrt{13}}{3}\right)^2 = \pi \cdot \frac{4 \cdot 13}{9} = \frac{52\pi}{9} \]

### Final Answer:
The area of the locus of the point at which the two circles subtend equal angles is \(\frac{52\pi}{9}\) square units.