Solution to eye-to-eye problem posted on Futility Closet

Futility Closet posted a Mathematical observation in Circles and Straightlines named “eye-to-eye” on May 1st, 2011. I wanted to prove this observation. Due to some busy schedule and travel I couldnt do this. When I met my father a week ago, I discussed the problem with him and today, found some time to close this. This post is a record of the solution.

Futility Closet‘s Greg Ross posts some very interesting Mathematical Observations from time to time. It is as fun to read as it is to work out and solve the problems. I tried solving one such problem before, but have been unsuccessful so far. But with, eye-to-eye, I have a solution.

The observation was posted in the form of a figure (below)

From Eye-to-Eye posted on Futility Closet

Image Courtesy: Futility Closet

The observation is

Draw two circles of any size and bracket them with tangents, as shown.

The chords in green will always be equal.

Heres the proof of this observation:

Choose the center of one of the circles to be horizon (0,0) such that the other circle’s center falls on +ve x-axis and line joining the two centers is x-axis. This means, we have two circles. As shown in this figure

Choice of Catesian Horizon for eye-to-eye

Figure describes the choice of Cartesian co-ordinates.

C0  ≡ x² + y² = r0² and C1 ≡ (x-x1)² + y² = r1²

We assume x1 > C0.

Let tangent to Cicle C0 be the line

y = m1x + p1 —————– (1)

If straight line (1) is a tangent that passes through (x1,0) – center of Circle C1, then,

0 = m1(x1) + p1

So, p1 = -m1x1.

Thus, tangent to Circle C0 that passes through center of circle C1 is

y = m1(x-x1) ——————- (2)

Let tangent to Circle C1 be the line

y = m0x + p0 ——————– (3)

If straight line (2) is a tangent that passes through (0,0) – center of Circle C0, then, equation of the tangent to Circle C1 that passes through (0,0) should be

y = m0x ————— (4)

Applying condition of tangency to straight line (2) and Circle C0,

r0² (1 + m1²) = p1²

r0² (1 + m1²) = m1²x1²

m1² = r0²/(x1²-r0²) —————– (5)

Similarly applying condition of tangency to straight line (4) and Circle C1,

r1² (1 + m0²) = (0 – m0x1 – 0)²

r1² (1 + m0²) = m0²x1²

m0² = r1²/(x1²-r1²) ——————— (6)

Straighline (2) intersects Circle C1 at 4 points.

Solving the two equations

y = m1(x-x1)

and

(x-x1)² + y² = r1²

We get,

(x-x1)² + m1²(x-x1)² = r1²

(1 + m1²)(x-x1)² = r1²

Substituting value of m1² from (5),

(x-x1)² (x1²)/(x1²-r0²) = r1²

Thus

x = x1 + (r1√(x1²-r0²))/x1

since x > 0 as explained earlier.

Solving for y, we get

y = m1 (m1√(x1²-r0²))

Substituting value of m1 from (5),

y = ±(r0r1√(x1²-r0²))/(x1√(x1²-r0²))

Thus

y = ±r0r1/x1 ——————– (7)

Straighline (4) intersects Circle C0 at 4 points.

Solving

y = m0x

and

x² + y² = r0²

We get,

x²(1 + m0²) = r0²

Substituting value of mo² from (6),

x² (x1²/(x1²-r1²)) = r0²

Thus

x = (r1√(x1²-r1²))/x1

since x > 0 as explained earlier.

Solving for y,

y = m0 (r0√(x1²-r1²))/x1

Substituting value of m0 from (6),

y = ±(r0r1√(x1²-r1²))/x1(√(x1²-r1²))

Thus

y = ±r0r1/x1 ————- (7)

So the two points of intersections of Straight line (2) on C1 that form a chord in C0 are

A1: (x1 + (r1√(x1²-r0²))/x1,r0r1/x1)

and

A1: (x1 + (r1√(x1²-r0²))/x1,-r0r1/x1)

And the two points of intersection of Straight line (4) on C0 that form a chord in C1 are

B1: ((r1√(x1²-r1²))/x1,r0r1/x1)

and

B2: ((r1√(x1²-r1²))/x1,-r0r1/x1)

Thus the two chords

A1A2 and B1B2 are of the same length,

Length of each chord is

(r0r1/x1)√2

Share and Enjoy, Sirius Cybernetics Style 😀

Update: Some formatting changes for improving readability of equations.

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