## Solution.

Let $T(x, y)$ be a point on the desired curve. The given condition $d(A, T)=2 d(B, T)$ can be written as

$$
\sqrt{(x+3)^{2}+(y-6)^{2}}=2 \sqrt{(x-6)^{2}+(y-3)^{2}}
$$

which, after squaring,

$$
(x+3)^{2}+(y-6)^{2}=4\left((x-6)^{2}+(y-3)^{2}\right)
$$

and simplifying,

$$
x^{2}+y^{2}-18 x-4 y+45=0
$$

becomes $\quad(x-9)^{2}+(y-2)^{2}=40$.

(5 points)

![](https://cdn.mathpix.com/cropped/2024_05_30_70b7ed256f659785a3e5g-4.jpg?height=546&width=818&top_left_y=264&top_left_x=643)

(5 points)

The desired set of points is a circle with center at point $S(9,2)$ and radius $r=2 \sqrt{10}$.

The intersections with the $x$-axis are obtained by solving the system

$$
\left\{\begin{array}{c}
(x-9)^{2}+(y-2)^{2}=40 \\
y=0
\end{array}\right.
$$

These are the points $(3,0)$ and $(15,0)$.

The circle does not intersect the $y$-axis, as the distance from the center to the $y$-axis is 9, which is greater than the radius of the circle. (This can also be concluded by solving the system.)

(2 points)
3. Given points $A(-3,6)$ and $B(6,3)$. Determine the set of all points whose distance to point $A$ is twice the distance to point $B$. Draw the resulting curve in the coordinate system and determine its intersections with the coordinate axes.