The center of a circle is $(4, 0)$ and the radius $4$. Inside the circle along the radius lies a point $P(x, y)$, through this point we draw a chord such that it's perpendicular towards the radius. What curve does the points $P(x, y)$ form for which the distance $|PQ|$ is equal to the distance between point $P$ and $(2, 0)$?
I tried to translate the problem from my native language to English so apologies for the bad translation. Hopefully the picture helps.
So the equation for the circle is $(x-4)^2+(y-0)^2=4^2$ and the length of $P$ and $(2, 0)$ turned out to be $\sqrt{x^2-4x+4+y^2}$.
Also the length between $P$ and $Q$ is $|PQ| = \sqrt{(x_2-x_1)^2+(y_2 -y_1)^2}$, but I don't really see how any of these help here at all. What should i do here?
Best Answer
Hint:
You’ve got the first equation correct. Now notice that if you connect the center of the circle and $Q$, you get a right triangle. Now compute the distance $OP$ ($O$ is the center), and use Pythagoras to calculate $PQ$. Then equate $PQ$ to the formula you obtained earlier and you’ll get the locus of $P$.