*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)

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

C

_{0}≡ x² + y² = r_{0}² and C_{1}≡ (x-x1)² + y² = r_{1}²

We assume x_{1} > C_{0}.

Let tangent to Cicle C_{0} be the line

y = m

_{1}x + p_{1}—————– (1)

If straight line (1) is a tangent that passes through (x_{1},0) – center of Circle C_{1}, then,

0 = m

_{1}(x_{1}) + p1So, p1 = -m

_{1}x_{1}.

Thus, tangent to Circle C_{0} that passes through center of circle C_{1} is

y = m

_{1}(x-x_{1}) ——————- (2)

Let tangent to Circle C1 be the line

y = m

_{0}x + p_{0}——————– (3)

If straight line (2) is a tangent that passes through (0,0) – center of Circle C_{0}, then, equation of the tangent to Circle C_{1} that passes through (0,0) should be

y = m

_{0}x ————— (4)

Applying condition of tangency to straight line (2) and Circle C_{0},

r

_{0}² (1 + m_{1}²) = p_{1}²r

_{0}² (1 + m_{1}²) = m_{1}²x_{1}²m

_{1}² = r_{0}²/(x_{1}²-r_{0}²) —————– (5)

Similarly applying condition of tangency to straight line (4) and Circle C_{1},

r

_{1}² (1 + m_{0}²) = (0 – m_{0}x_{1}– 0)²r

_{1}² (1 + m_{0}²) = m_{0}²x_{1}²m

_{0}² = r_{1}²/(x_{1}²-r_{1}²) ——————— (6)

Straighline (2) intersects Circle C_{1} at 4 points.

Solving the two equations

y = m

_{1}(x-x_{1})

and

(x-x

_{1})² + y² = r_{1}²

We get,

(x-x

_{1})² + m_{1}²(x-x_{1})² = r_{1}²(1 + m

_{1}²)(x-x_{1})² = r_{1}²

Substituting value of m_{1}² from (5),

(x-x

_{1})² (x_{1}²)/(x_{1}²-r_{0}²) = r_{1}²

Thus

x = x

_{1}+ (r_{1}√(x_{1}²-r_{0}²))/x_{1}

since x > 0 as explained earlier.

Solving for y, we get

y = m

_{1}(m_{1}√(x_{1}²-r_{0}²))

Substituting value of m1 from (5),

y = ±(r

_{0}r_{1}~~√(x~~)/(x1_{1}²-r_{0}²)~~√(x~~)_{1}²-r_{0}²)

Thus

y = ±r

_{0}r_{1}/x_{1}——————– (7)

Straighline (4) intersects Circle C_{0} at 4 points.

Solving

y = m

_{0}x

and

x² + y² = r

_{0}²

We get,

x²(1 + m

_{0}²) = r_{0}²

Substituting value of mo² from (6),

x² (x

_{1}²/(x_{1}²-r_{1}²)) = r_{0}²

Thus

x = (r

_{1}√(x_{1}²-_{r1}²))/x_{1}

since x > 0 as explained earlier.

Solving for y,

y = m

_{0}(r_{0}√(x_{1}²-r_{1}²))/x_{1}

Substituting value of m0 from (6),

y = ±(r

_{0}r_{1}~~√(x~~)/x_{1}²-r_{1}²)_{1}(~~√(x~~)_{1}²-r_{1}²)

Thus

y = ±r

_{0}r_{1}/x_{1 }————- (7)

So the two points of intersections of Straight line (2) on C_{1} that form a chord in C_{0} are

A

_{1}: (x_{1}+ (r_{1}√(x_{1}²-r_{0}²))/x_{1,}r_{0}r_{1}/x_{1})

and

A

_{1}: (x_{1}+ (r_{1}√(x_{1}²-r_{0}²))/x_{1,}-r_{0}r_{1}/x_{1})

And the two points of intersection of Straight line (4) on C_{0} that form a chord in C_{1} are

B

_{1}: ((r_{1}√(x_{1}²-_{r1}²))/x_{1},r_{0}r_{1}/x_{1})

and

B

_{2}: ((r_{1}√(x_{1}²-_{r1}²))/x_{1},-r_{0}r_{1}/x_{1})

Thus the two chords

A

_{1}A_{2}and B_{1}B_{2}are of the same length,

Length of each chord is

(r

_{0}r_{1}/x_{1})√2

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**Update**: Some formatting changes for improving readability of equations.