# Two Circles and a Limit Trigonometric Solution

We are solving the Two Circles and a Limit problem:

A stationary circle of radius 3 is centered at (3, 0). Another circle of variable radius r is centered at the origin and meets the positive y-axis in point A. Let B be the common point of the two circles in the upper half-plane. Let E be the intersection of AB extended with the x-axis. What happens to E as r grows smaller and smaller?

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

The use of trigonometry was suggested by the NY math teacher Patrick Honner.

Let α = ∠AOB, with O standing for the origin. Then B = (r sinα, r cosα), so that the slope of AB is given by

(r cosα - r) / (r sinα) = - sin(α/2)/cos(α/2) = -tan(α/2) = tan(180° - α/2).

This means that ∠AEO = α/2. In the right ΔAOE, EO = r / tan(α/2).

But r is a function of α - r = r(α) - which can be determined from the stationary circle. Join B to F, the second end of the diameter. ΔOBF is right, OF = 6, and ∠OFB = α. It follows that r = OB = 6 sin(α).

Of course we also have

EO = r / tan(α/2) = 6 sin(α) / tan(α/2) = 12 cos²(α/2).

Clearly α and r tend to 0 simultaneously. So, as r tends to 0, cos²(α/2) tends to 1, while EO has the limit of 12.

### Note

Patrick Honner came up with a different opening for the proof.

Let X be the center of the stationary circle. ΔABX is isosceles and OX = BX = 3, OB = r. Relative to that circle, angle AOB is bounded by a tangent and a chord, OB, that subtends the central angle OXB. Therefore, ∠OXB = 2α. Dropping a perpendicular from X onto OB, we immediately obtain OB/2 = 3sin(α), or r = 6 sin(α).