Consider I have two points p and q, and a line segment l: y=mx+c (actually the enpoints of the segment are given). There is a circle with center q which is growing with time t, i.e. the radius r = k.t where k is some constant. Consider z(t) be a point(s) of intersection between the line segment and the growing circle.

What would be the shape of the graph between d(p, z(t)) and t, where d(p, z(t)) is the distance between point p and z(t) . we take the intersection point z(t) which is far from p.

the picture shows growing circle with time and intersection point *z(t)* with the line segment. thin line is the distance *d(p, z(t))*.

I can find the intersection points z(t) at any time t because the radius and the center are known. Then I simply calculate the distance between p and z(t). I get the intuition that the graph will be similar to the dark line shown in picture below. Is it possible that the curve can go above the dashed line.

enter image description here

  • $\begingroup$ its a more complex problem. this is just a part of that. I want to know the increase in distance between p and z(t) with time t. $\endgroup$ – CODError Jul 21 '16 at 9:03
  • $\begingroup$ That doesn't answer my question. What have you tried? $\endgroup$ – Jossie Calderon Jul 21 '16 at 9:06
  • $\begingroup$ DO you mean my approach to this? $\endgroup$ – CODError Jul 21 '16 at 9:07
  • $\begingroup$ Yes @CODError.. $\endgroup$ – Jossie Calderon Jul 21 '16 at 9:12
  • $\begingroup$ I have put some extra details. I couldnt think of a proper approach. I am stuck with the details I have provided. Now, my main concern is to prove the the dark line curve could not go above dashed line shown in the example graph. I am actually doing a linear interpolation, and I dont want the actual distance value to have higher value that the linearly interpolated value at any time t $\endgroup$ – CODError Jul 21 '16 at 9:28

Use coordinate geometry. The line L is along the $x$-axis, the point $P(0,p)$ is on the $y$-axis, $Q(a,b)$ is the second point and the radius $r= kt$. The equation of the circle $$(x-a)^2+(y-b)^2=k^2t^2$$ Put $y=0$ and solve for $x$ to get $$z(t)=a+\sqrt{k^2t^2-b^2}$$ and the distance $d$ is given by $$d^2=p^2+\left(a+\sqrt{k^2t^2-b^2}\right)^2$$ Special case: $p=5, a=4, b=3, k=1$


This curve is above the dotted line for $3\leq t \leq 10$


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