# Thread: Drawing lines of equal length

1. ## Drawing lines of equal length

Here is another problem I can't seem to solve.

How do you make those two segments(which is the radii of the blue circle) equal when you are given the small circle's center and the big circle's center and ruler and compass?
(It only shows half of a circle, though) (also, the blue circle is touching the small circle externally and touching the big circle internally)
When I extend the two segments, one of them meets the small circle's center and other other meets the big circle's center.

2. Let AB be the diameter of larger circle and CD be the diameter of the smaller circle. Let P be the mid point of AC and Q be the midpoint of DB.
Let O be the mid point of PQ. Taking O as the center and OP as the radius draw a circle. Every point on this circle is equidistant from the larger and the smaller circle. Draw a radius from the larger circle. Point of intersection of the radius and in between circle will serve as the center.

3. Thanks a lot. But that doesn't work either. That doesn't always make those two segments equal.
Are there more accurate way to do it?
And by the way, that shape with those dotted lines, it's not a circle. It's an ellipse.

4. Draw a smaller circle of center O1 and radius R1.
Draw a larger circle of center O2 and radius R2.
Let the distance between O1 and O2 be d. Let O be the mid point of O1O2.
Let AB be the diameter of smaller circle and CD be the diameter of the larger circle.
Find the mid point P of BD.
Now OP is the is the semi-major axis a of the ellipse.
Semi-minor axis b is given by
b = sqrt[a^2 - (d/2)^2]
Any point Q on the ellipse is given by ( a*cosθ, b*sinθ),where θ is the angle between OQ and OP.
Join O1Q which cuts the smaller circle at R. O1Q is along the radius of the smaller circle. Hence it is normal to the smaller circle.
Join O2Q. Extend it to meet the larger circle at S. OQ is along the radius of the larger circle. Hence it is normal to the larger circle.
The point of intersection of O1Q and O2Q will be the center of the required circle.
Hence QR = QS.