FORM GR1268 Question 10

Question:
\[\sqrt{(x+3)^2+(y-2)^2}=\sqrt{(x-3)^2+y^2}\]
In the \(xy\)-plane, the set of points whose coordinates satisfy the equation above is
(A) a line      (B) a circle     (C) an ellipse     (D) a parabola     (E) one branch of a hyperbola

Answer:
(A)

Answer Key:
Square both sides:
\[\begin{align}
(x+3)^2+(y-2)^2&=(x-3)^2+y^2\\
x^2+6x+9+y^2-4y+4&=x^2-6x+9+y^2\\
12x-4y+4=0
\end{align}\]
This is a line.


Alternatively, we can think this problem geometrically.

\(\sqrt{(x+3)^2+(y-2)^2}=r\) is a circle with radius \(r\) centered at \((-3,2)\).

\(\sqrt{(x-3)^2+y^2}=r\) is a circle with radius \(r\) centered at \((3,0)\).

The distance between the centers is \(\sqrt{(-3-3)^2+(2-0)^2}=\sqrt{40}=2\sqrt{10}\).

When \(r\lt\sqrt{10}\), the circles do not intersect.

When \(r=\sqrt{10}\), the circles intersect at one point.

When \(r\gt\sqrt{10}\), the circles intersect at two points.

The collection of the intersections is a line that is equidistant from the centers of the two circles.