Problem β-4: Find a tangent to a circle at a point.

Solution

1. Draw a circle with center at O and the given point p.

   

2. Draw an arbitrary point q on the circle.

   

3. Draw a circle centered on q through p,

   

4. This defines a second intersection r between the circles. Draw a circle centered on p through r.

   

5. This intersects the second circle at s. Finally, the line ps is the desired tangent to the first circle at p.

   

Proof

Draw the radius Op of the first circle and the lines pr, pq, ps, qr, qs, rs and the angles α (pOq), β (prq), γ (qpr), δ (qps), ε (rpO), and φ (Oqp).

As p and q are on a circle centered at O, and r is another point on the same circle, then β=α/2. As p and r are also in a circle centered at q, the triangle pqr is isosceles. Therefore, γ=β=α/2. The sides of triangle pqs are the same as those of rqp; both triangles are congruent. Thus, δ=γ=α/2. The triangle pOq is isosceles, and therefore, the angle φ =ε+γ is (π-α)/2 and ε=π/2-α. Finally, the angle between the radius Op and the line ps is ε+γ+δ=(π/2-α)+(α/2)+(α/2)=π/2. As ps is orthogonal to the radius Op, it is tangent at p to the circle of radius Op, qed.