# Math Help - locus word problem

1. ## locus word problem

A dog walking through a yard. Tree A(-2,3), Tree B(4, -6).

Dog is at all times, twice as far from A as it from B.
Draw a graph showing the trees, path of dog and relationship defining the locus.

Write geometric description of the path of the dog, relative to the two trees.

I had no idea where to start, so I plotted the points on the graph and did the equation for Midpoint of AB and plotted that point - but it looks SO OFF!

$\frac {-2+4}{2} , \frac{3+(-6)}{2}
\\=1,4.5$

2. Originally Posted by sux@math
A dog walking through a yard. Tree A(-2,3), Tree B(4, -6).

Dog is at all times, twice as far from A as it from B.
Draw a graph showing the trees, path of dog and relationship defining the locus.

Write geometric description of the path of the dog, relative to the two trees.

I had no idea where to start, so I plotted the points on the graph and did the equation for Midpoint of AB and plotted that point - but it looks SO OFF!

$\frac {-2+4}{2} , \frac{3+(-6)}{2}
\\=1,4.5" alt="\frac {-2+4}{2} , \frac{3+(-6)}{2}
\\=1,4.5" />
Let a general point on the locus of the dog be P(x, y). Then dPA = 2 dPB.

So get the distance dPA between P and A and the distance dPB between P and B. Then substitute .....

By the way, the locus is a Circle of Apollonius. Some sites that might interest:

Circles of Apollonius

The circle of Apollonius

3. Hello, sux@math!

A dog walking through a yard. Tree $A(-2,3)$, Tree $B (4, -6)$

Dog is at all times, twice as far from $A$ as it is from $B.$

Draw a graph showing the trees, path of dog and relationship defining the locus.

Write geometric description of the path of the dog, relative to the two trees.
Code:
A   |             D
* - + - - - - - - * (x,y)
(-2,3) |            /
--------+-----------/--------
|          /
|         /
|        /
|       * (4,-6)
|       B

The trees are at $A\text{ and }B$; the dog is at $D(x,y)$

Since $DA \:=\:2\!\cdot\!DB$, we have: . $\sqrt{(x+2)^2+(y-3)^2} \;=\;2\cdot\sqrt{(x-4)^2+(y+6)^2}$

Square both sides: . $x^2 + 4x + 4 + y^2 - 6y + 9 \;\;=\;\;4(x^2-8x+16 + y^2 + 12y + 36)$

. . which simplifies to: . $x^2 - 12x + y^2 + 18y \;=\;65$

Complete the square: . $(x^2-12x \;{\color{blue}+\: 36}) + (y^2 + 18y\;{\color{red}+\: 81}) \;=\;65 \;{\color{blue}+\: 36}\:{\color{red} +\: 81}$

. . and we have: . $(x-6)^2 + (y+9)^2 \;=\;182$

The dog's path is a circle with center $(6,\:\text{-}9)$ and radius $\sqrt{182}$

4. You really are a *super* member, Soroban! A million, trillion thanks!!!!

5. ...further more, it is UNBELIEVABLY helpful to a visual learner like me that you put the different colours in your equation.

Honestly, thanks so much for taking the time - you have no idea how much you've helped me with your efforts!

6. Originally Posted by sux@math
...further more, it is UNBELIEVABLY helpful to a visual learner like me that you put the different colours in your equation.

Honestly, thanks so much for taking the time - you have no idea how much you've helped me with your efforts!
Did you try doing the problem using:
1. The suggestions I gave?
2. Taking a look at the links I provided and playing around with the dynamic geometry .....

*Sigh* If only super members could be brought into exams like calculators can be. Alas .....

7. Haha! Mr. Fantastic! I am so sorry if I offended you! To a math n00b like me, your answer seemed either a) glib or b) waaayyyyy over my head and the link gave me a small seizure, or c) both.

Yes, I did try your suggestion but had no luck until I saw Soroban's post.

I apologize from the bottom of my cold-black heart!