Problem:
Find the locus of a point the sum of the squares of whose distances from $(2,3)$ and $(-1,-2)$ is $34$.
Solution:
Let $(x,y)$ be the point. Then $(x-2)^2+(y-3)^2+(x+1)^2+(y+2)^2=34$. Simplify: $x^2+y^2-x-y-8=0$, the equation of a circle.
Source: Schaum's 3000 Solved Problems in Calculus
I read that locus is the set all points the share a property. So is locus of a point necessarily a circle? How does mentioning locus of a point affect the answer? Also, is the problem worded incorrectly? It sounds grammatically incorrect. Finally, I am having trouble seeing how they set up the equation and also having difficulty picturing what is going on. I need a better explanation.
In general, the locus need not be a circle. But the definition of the locus is important: it is the set of all points which share a certain property. In this case the property is that the sum of the squares of the distances from all points that form the locus to two given points is a constant.
To visualize the solution, draw a cartesian plane. Set point $A$ at coordinates $(2,3)$ and set point $B$ at coordinates $(-1,-2)$. Now mark a point $X$ at some coordinate $(x,y)$, and connect $X$ to $A$ and $X$ to $B$. Label these segments $d_1$ and $d_2$. The problem asks what are all possible coordinates of $X$ such that $d_1^2+d_2^2=34$. So, we write $$ d_1^2 = (x-2)^2 + (y-3)^2 \\ d_2^2 = (x+1)^2 + (x+2)^2 $$ which plugged into the condition yields, after simplification, $$x^2+y^2-x-y=8.$$
What does this equation describe? We take one more step and write it as $$\left(x-\frac{1}{2}\right)^2 + \left(y-\frac{1}{2}\right)^2 = 8.5$$
But this clearly is a circle centered at point $(1/2,1/2)$ with radius $\sqrt{8.5}$.
Thus, any point from the circumference of this circle would satisfy the property that the sum of the square of the two distances equals 34. Here's how the diagram would look:
Notice that segment $AB$ is in fact a diameter of this circle, thus $\angle{AXB}=90^{\circ}$ (it spans half a circle). Therefore, triangle $ABX$ is rectangular, and by Pythagora's, $d_1^2 + d_2^2 = AB^2$. But $AB$ is the diameter, thus $AB^2 = (2r)^2 = 4r^2 = 4(8.5) = 34$.