0

Given point $A$ on the circle $x^2+y^2=R^2$.

From $A$ passes parallel line to the x-axis.

On this parallel line we Assign from point $A$

a Section with Length $2R$ at the Positive direction of the x-axis and we get the point $P(x,y)$.

How can i prove that $p(x,y)$ is a circle with radius $R$ and center $(2R,0)$,

And if the centers of all circles which tangent to the circle $P(x,y)$ and to the line $x=-t$ is a Canonical parabola so what is the parabola equation and $t$ by $R$?

I tried to sketch but it went wrong.

bori12
  • 65
  • Your description of $P(x,y)$ seems to be just a very detailed way of saying "take any point $A$ on the original circle and translate it $2R$ units to the right (i.e., in the direction of the positive $x$-axis). Since translation preserves distances and therefore respects circles, radii, and centers, the locus of these points $P(x,y)$ is a circle of the same radius as the original one with its center translated $2R$ units to the right. – Andreas Blass Jun 24 '13 at 01:03
  • @AndreasBlass - that's right. i guess the main problem is second part; to show that $t=R$ and $y^2=8RX$ – bori12 Jun 24 '13 at 04:36

1 Answers1

1

  1. By construction, point $(x,y)$ belongs to the new set if and only if $(x-2R,y)$ lies on the circle $x^2+y^2=R^2$. This yields the equation $(x-2R)^2+y^2=R^2.

  2. Let $C$ be the circle described above. A circle with center $(x,y)$ and radius $r$ is (externally) tangent to $C$ if and only if the distance from $(x,y)$ to $(2R,0)$ is $r+R$. For this circle to be also tangent to vertical line $X=-t$, we need $x+t=r$. Thus, the distance from $(x,y)$ to $(2R,0)$ is $x+t+R$. From $$(x-2R)^2+y^2=(x+t+R)^2$$ we obtain $$2x(3R+t) = y^2+4R^2-(t+R)^2$$ which is an equation of parabola. I leave it for you to determine what combination of parameters makes this parabola "canonical".

40 votes
  • 9,736