Task B-4.5. Determine the equation of the curve along which point $A$ travels, if its distance from the origin is always half the distance from point $T(3,6)$.
## First solution.

Let $A(x, y)$ and $O(0,0)$. Then

$$
\begin{aligned}
& 2|\overline{A O}|=|\overline{A T}| \\
& 2 \sqrt{x^{2}+y^{2}}=\sqrt{(x-3)^{2}+(y-6)^{2}} \\
& 4\left(x^{2}+y^{2}\right)=x^{2}-6 x+9+y^{2}-12 y+36 \\
& 3 x^{2}+3 y^{2}+6 x+12 y-45=0
\end{aligned}
$$

(1 point)

Divide the given equation by 3. Then

$$
\begin{aligned}
& x^{2}+y^{2}+2 x+4 y-15=0 \\
& x^{2}+2 x+1+y^{2}+4 y+4-15-1-4=0 \\
& (x+1)^{2}+(y+2)^{2}=20
\end{aligned}
$$

Thus, the point $A$ travels on the circle $(x+1)^{2}+(y+2)^{2}=20$.