The perimeter of the locus of the point at which the two circles $x^2+y^2=1$ and ${\left(x-4\right)}^2+y^2=4$ subtend equal angles is

• A

  $\left(\frac{4}{3}\right)\pi$
• B

  $\left(\frac{8}{3}\right)\pi$
• C

  $\left(\frac{2}{3}\right)\pi$
• Correct answer
  D

  $\left(\frac{16}{3}\right)\pi$

Solution

$\sin \theta =\frac{1}{\sqrt{h^2+k^2}}=\frac{2}{\sqrt{{\left(h-4\right)}^2+k^2}}$
$\Rightarrow h^2-8h+16+k^2=4h^2+4k^2$
$\Rightarrow 3x^2+3y^2+8x-16=0$
Radius of the circle is $\sqrt{{\left(\frac{4}{3}\right)}^2+\frac{16}{3}}=\sqrt{\frac{16+48}{9}}$
$=\frac{8}{3}$
⇒ Perimeter is equal to $\frac{16\pi }{3}$

Hence, the correct answer is option: 'D'.