The perimeter of the locus of the point at which the two circules `x^(2)+y^(2)=1 and (x-4)^(2)+y^(2)=4` subtend equal angles is

Options

A. `((4)/(3))pi`
B. `((8)/(3))pi`
C. `((2)/(3))pi`
D. `((16)/(3))pi`

Answer

To find the perimeter 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, we can follow these steps:

### Step 1: Identify the Circles
The first circle is given by:
\[ x^2 + y^2 = 1 \]
- Center: \( (0, 0) \)
- Radius: \( 1 \)

The second circle is given by:
\[ (x - 4)^2 + y^2 = 4 \]
- Center: \( (4, 0) \)
- Radius: \( 2 \)

### Step 2: Use the Angle Subtended Condition
To find the locus of points where the two circles subtend equal angles, we can use the property of angles subtended by chords. The sine of the angle subtended by a point \( (h, k) \) from the centers of the circles can be expressed as:
\[ \sin \theta_1 = \frac{1}{\sqrt{h^2 + k^2}} \]
for the first circle, and
\[ \sin \theta_2 = \frac{2}{\sqrt{(h - 4)^2 + k^2}} \]
for the second circle.

### Step 3: Set the Sines Equal
Since the angles are equal, we set the sines equal:
\[ \frac{1}{\sqrt{h^2 + k^2}} = \frac{2}{\sqrt{(h - 4)^2 + k^2}} \]

### Step 4: Cross Multiply and Simplify
Cross multiplying gives:
\[ \sqrt{(h - 4)^2 + k^2} = 2\sqrt{h^2 + k^2} \]

Squaring both sides results in:
\[ (h - 4)^2 + k^2 = 4(h^2 + k^2) \]

Expanding both sides:
\[ h^2 - 8h + 16 + k^2 = 4h^2 + 4k^2 \]

### Step 5: Rearranging Terms
Rearranging gives:
\[ 0 = 3h^2 + 3k^2 + 8h - 16 \]

### Step 6: Replace Variables
Let \( h = x \) and \( k = y \):
\[ 3x^2 + 3y^2 + 8x - 16 = 0 \]

### Step 7: Simplify to Standard Form
Dividing through by 3 gives:
\[ x^2 + y^2 + \frac{8}{3}x - \frac{16}{3} = 0 \]

### Step 8: Complete the Square
To write this in standard form, complete the square for \( x \):
\[ x^2 + \frac{8}{3}x + y^2 = \frac{16}{3} \]

Completing the square for \( x \):
\[ \left(x + \frac{4}{3}\right)^2 - \frac{16}{9} + y^2 = \frac{16}{3} \]

This simplifies to:
\[ \left(x + \frac{4}{3}\right)^2 + y^2 = \frac{16}{3} + \frac{16}{9} \]
\[ = \frac{48}{9} + \frac{16}{9} = \frac{64}{9} \]

### Step 9: Identify the Circle's Center and Radius
The standard form of the circle is:
\[ \left(x + \frac{4}{3}\right)^2 + y^2 = \left(\frac{8}{3}\right)^2 \]
- Center: \( \left(-\frac{4}{3}, 0\right) \)
- Radius: \( \frac{8}{3} \)

### Step 10: Calculate the Perimeter
The perimeter \( P \) of a circle is given by:
\[ P = 2\pi r \]
Substituting the radius:
\[ P = 2\pi \left(\frac{8}{3}\right) = \frac{16\pi}{3} \]

### Final Answer
Thus, the perimeter of the locus is:
\[ \frac{16\pi}{3} \]

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