Hey guys,
I need to find the angle that is in the picture below. I know the radius of the both cirlces (they are both the same) and I know the distance between the two circles.
I am baffled as to how I should approach this!
Hey guys,
I need to find the angle that is in the picture below. I know the radius of the both cirlces (they are both the same) and I know the distance between the two circles.
I am baffled as to how I should approach this!
I'm always a fan of Brute Force.
Put your circles on a set of coordinate axes, once circle (radius r and center (0,0)) at the origin and the other (radius R and center (a,0)) on the positive x-axis.
Points on the origin circle look like this: (b,c) where $\displaystyle b^{2} + c^{2} = r^{2}$
Points on the remote circle look like this: (d,e) where $\displaystyle (d-a)^{2} + e^{2} = R^{2}$
Just to emphasize that the circles do not intersect, let's also define f > 0 so that a = r + f + R
The line has equation $\displaystyle y-e = \frac{e-c}{d-b}(x-d)$
You can use calculus or or algebra to show that the line is perpendicular to a radius at the points of tangency.
$\displaystyle \frac{e-c}{d-b} = -\frac{b}{c}$ and $\displaystyle \frac{e-c}{d-b} = -\frac{d}{e}$
That's enough information to find the points. The angles aren't too tricky after that. Let's see what you get. It will be fun!
Note: I have no doubt there is an elegant way to do this, like tocbol's geometry solution. I couldn't do that because I did not use the fact that the radii were the same. My solution is more general. If nothing else, this little demonstration should encourage you that you ALWAYS should be able to find SOME approach. It may be ugly, but it's better than the frustration inherent in just staring at it. Once you have the solution, you can search for elegance.
Whether the centers of the two circles are in the same horizontal line or not, the solution will be the same.
The common tangent line is perpendicular to the radii of both circles at the points of tangency.
Given or known:
>>>radius of both circles is, say, r.
>>>distance between the two centers is, say, d.
Draw the d.
The slant line or the common tangent line bisects the d.
Two similar and congruent right triangles are formed.
Each one has a leg of r and a hypotenuse of d/2.
The angle we are solving for, the central angle, is included in these two known sides.
So,
cos(theta) = r / (d/2)
cos(theta) = 2r/d
Therefore,
theta = arccos(2r/d) ---------answer.
ok i assume that the angle u are referring to is the angle formed between the two radii.
u also said that the following is known:
- the radius of both circles, which u said are the same and that implies that the shape of both circles are the same ie. same area, circumference etc.
- the distance between the two circles
So what u need to do is to complete the diagram 1st ..... connect the distance between the 2 circles via a straight line to their radii. (Also the tangent would cut this line (the distance between the 2 circles) in half, so they will be equidistance apart.)
Therefore the angles that are required to find, now become alternate angles which are equal in size.
Furthermore, we know the theorem which states that a radius is perpendicular to a tangent at the point of contact.
Based on the above info u should now be able to form a triangle, whose 2 sides are known (one of the radius and the other radius plus half the distance between the 2 circles) as well as the perpendicular angle (90 degree).
Thus using the sine rule $\displaystyle \frac{a}{sin A} = \frac{b}{sin B}$, substitute the 3 values above and hence find the size of the required angle.
Hello, VooDoo!
I need to find the angle that is in the picture below.
I know the radius of the both cirlces (they are both the same)
and I know the distance between the two circles.Code:o * B * θ | o* | * o |r C E * o | o o * * * * * * * * * * * o o | o * D r| o * | *o |θ * A * o
The circles have centers at A and B.
The radius is: .$\displaystyle AC \,=\, BD \,=\, r$
The line joining the centers is $\displaystyle AB \,= \,d$
The common tangent is $\displaystyle CD$
They intersect at $\displaystyle E.$
Right triangles ACE and BDE are congruent.
. . Hence: .$\displaystyle AE \,=\,EB \,=\,\frac{1}{2}d$ .and .$\displaystyle \theta \,=\,\angle CAE \,=\,\angle DBE$
In right triangle $\displaystyle ACE\!:\;\;\cos\theta \:=\:\frac{AC}{AE} \:=\:\frac{r}{\frac{1}{2}d}$
Therefore: .$\displaystyle \theta \;=\;\arccos\left(\frac{2r}{d}\right)$
Ahhh, ticbol beat me to it!
.