Finding parallel segments between two circles

[BorisGrushenko]

Hello folks!

I have an interesting geometry task which i am trying to solve.

Well, both circles with centers are given. Further the segment a is given too. I have to find all parallel segments to a with same length as a, such that one end of these segments lies on Circle 1 and the other end lies on Circle 2.


That looks kind of tricky to get to me. I need some kind of hint how to do it. Some suggestions would be nice.

Thanks!
Boris

 

[Ishuda]

Hello folks!

I have an interesting geometry task which i am trying to solve.

Well, both circles with centers are given. Further the segment a is given too. I have to find all parallel segments to a with same length as a, such that one end of these segments lies on Circle 1 and the other end lies on Circle 2.


That looks kind of tricky to get to me. I need some kind of hint how to do it. Some suggestions would be nice.

Thanks!
Boris

You could start by writing the equations for the circle and line represented by the given line segment: As something to make life easier (IMO), call the center of the lower circle (0,0) so that the lower circle's equation is
x² + y² = r₀²
and that of the upper circle is
(x-x₁)² + (y-y₁)² = r₁².
The equation of the line with the given line segment a is
L_a: y = y_a + s (x-x_a)

Now put things together. Given any x(u)=u on the lower circle,
y(v) = v(u) = v = \(\displaystyle \pm \sqrt{r_0^2\, -\, u^2}\)
The line parallel to segment a going through point (u,v) [actually (u,v₁) for the plus and (u,v₂) for the minus)] is
y = v + s (x-u)
Parametrically this can be written as
x = u + \(\displaystyle \alpha\) t
y = v + \(\displaystyle \alpha\) s t
where
\(\displaystyle \alpha\, =\, \frac{1}{1+s^2}\)
and t is the distance along the line in the positive x direction, i.e. as t increases, x increases. So go a distance up (or down) the line and see if you are on the upper circle.

Now actually, that answers the question 'given a value of x on the lower circle, is the distance to the upper circle along a line parallel to segment a the same length of segment a'. However, I do think this might give some incite into solving the original question. Then again, maybe not.

 

[Deleted member 4993]

You could start by writing the equations for the circle and line represented by the given line segment: As something to make life easier (IMO), call the center of the lower circle (0,0) so that the lower circle's equation is
x² + y² = r₀²
and that of the upper circle is
(x-x₁)² + (y-y₁)² = r₁².
The equation of the line with the given line segment a is
L_a: y = y_a + s (x-x_a)

Now put things together. Given any x(u)=u on the lower circle,
y(v) = v(u) = v = \(\displaystyle \pm \sqrt{r_0^2\, -\, u^2}\)
The line parallel to segment a going through point (u,v) [actually (u,v₁) for the plus and (u,v₂) for the minus)] is
y = v + s (x-u)
Parametrically this can be written as
x = u + \(\displaystyle \alpha\) t
y = v + \(\displaystyle \alpha\) s t
where
\(\displaystyle \alpha\, =\, \frac{1}{1+s^2}\)
and t is the distance along the line in the positive x direction, i.e. as t increases, x increases. So go a distance up (or down) the line and see if you are on the upper circle.

Now actually, that answers the question 'given a value of x on the lower circle, is the distance to the upper circle along a line parallel to segment a the same length of segment a'. However, I do think this might give some incite into solving the original question. Then again, maybe not.

Are you trying to incite a revolt here??!

Or Microsoft auto-correct sent you off to this violence-filled state of mind?

 

[Ishuda]

Are you trying to incite a revolt here??!

Or Microsoft auto-correct sent you off to this violence-filled state of mind?

Just trying to make it more interesting.