Find Coordinates of a Point Based on Distances Relative to Two Other Points

Question

I have the following question:

Point $P(x,y)$ is $6$ units from $A (0,0)$ and $9$ units from $B(9,0)$. Find values of $x$ and $y$.

Do I need to incorporate the usage of the distance formula with this problem? If so, how? If not, then what other method(s) can I use to solve this problem?

Answer 1

$$x^2+y^2=36$$ and $$(x-9)^2+y^2=81.$$ Thus, $$(x-9)^2-x^2=81-36$$ or $$x=2$$ and ...

Answer 2

If you think geometrically, then all points that are $r_1=6$ units from $A(0,0)$ form a circle, and all points that are $r_2=9$ units from $B(9,0)$ form another circle — so you're looking for the point(s) of intersection of two circle. There are three options then:

• $\operatorname{dist}(A,B)>r_1+r_2$: the circles are too far from each other so that they don't intersect;
• $\operatorname{dist}(A,B)=r_1+r_2$: the circles are tangent to each other, thus intersecting only at a single point;
• $\operatorname{dist}(A,B)<r_1+r_2$: the circles intersect at two points.

Since here we have the last case, the circles intersect at two points, so this question has two solutions.

To find those solutions: yes, you should use the distance formula. Moreover, to simplify equations that we will need to solve, we can set up the squares of distances, thus eliminating the square root from our equations.

The first condition tells you that $$\operatorname{dist}(P,A)=\sqrt{(x-0)^2+(y-0)^2}=\sqrt{x^2+y^2}=6 \quad \Longrightarrow \quad x^2+y^2=36.$$ Similarly, the second condition will give you another equation. Solving this system of two equations for $x$ and $y$, you'll find the desired points.