Question

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 is

• A. $(\frac{4}{3})\pi$
• B. $(\frac{8}{3})\pi$
• C. $(\frac{2}{3})\pi$
• D. $(\frac{16}{3})\pi$

Answer 1

Answer: (D)

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

Answer 2

Let the two circles be $C_1: x^2 + y^2 = 1$ with center $O_1 = (0, 0)$ and radius $r_1 = 1$, and $C_2: (x - 4)^2 + y^2 = 4$ with center $O_2 = (4, 0)$ and radius $r_2 = 2$.

Let $P(x, y)$ be a point on the locus. The angle subtended by a circle of radius $r$ at an external point $P$ at a distance $d$ from the center is $2\alpha$, where $\sin \alpha = \frac{r}{d}$.

For $C_1$, the distance from $P(x, y)$ to $O_1(0, 0)$ is $d_1 = \sqrt{x^2 + y^2}$. The angle subtended is $2\alpha_1$, where $\sin \alpha_1 = \frac{r_1}{d_1} = \frac{1}{\sqrt{x^2 + y^2}}$. For $C_2$, the distance from $P(x, y)$ to $O_2(4, 0)$ is $d_2 = \sqrt{(x - 4)^2 + y^2}$. The angle subtended is $2\alpha_2$, where $\sin \alpha_2 = \frac{r_2}{d_2} = \frac{2}{\sqrt{(x - 4)^2 + y^2}}$.

The condition that the two circles subtend equal angles at $P$ implies $2\alpha_1 = 2\alpha_2$, so $\alpha_1 = \alpha_2$. Since $P$ must be external to both circles for the angles to be defined as above, $d_1 > r_1$ and $d_2 > r_2$, which means $0 < \sin \alpha_1 < 1$ and $0 < \sin \alpha_2 < 1$. Thus, $\alpha_1$ and $\alpha_2$ are acute angles.

Equating the sines: $\sin \alpha_1 = \sin \alpha_2$ $\frac{1}{d_1} = \frac{2}{d_2}$ $\frac{1}{\sqrt{x^2 + y^2}} = \frac{2}{\sqrt{(x - 4)^2 + y^2}}$ Squaring both sides: $\frac{1}{x^2 + y^2} = \frac{4}{(x - 4)^2 + y^2}$ $(x - 4)^2 + y^2 = 4(x^2 + y^2)$ Expanding and rearranging: $x^2 - 8x + 16 + y^2 = 4x^2 + 4y^2$ $3x^2 + 3y^2 + 8x - 16 = 0$ Dividing by 3: $x^2 + y^2 + \frac{8}{3}x - \frac{16}{3} = 0$ Completing the square for the $x$ terms: $\left(x^2 + \frac{8}{3}x\right) + y^2 = \frac{16}{3}$ $\left(x + \frac{4}{3}\right)^2 - \left(\frac{4}{3}\right)^2 + y^2 = \frac{16}{3}$ $\left(x + \frac{4}{3}\right)^2 + y^2 = \frac{16}{3} + \frac{16}{9}$ $\left(x + \frac{4}{3}\right)^2 + y^2 = \frac{48 + 16}{9}$ $\left(x + \frac{4}{3}\right)^2 + y^2 = \frac{64}{9}$ This is the equation of a circle with center $C = \left(-\frac{4}{3}, 0\right)$ and radius $R = \sqrt{\frac{64}{9}} = \frac{8}{3}$.

The perimeter of this locus circle is $2\pi R$. Perimeter $= 2\pi \left(\frac{8}{3}\right) = \frac{16\pi}{3}$.

The analysis in the raw solution confirms that all points on this locus are indeed external to both original circles, validating the method used.