# Euclid 1999 Question 4(a) - Circle Tangent Intersection

Below is a question and the intended solution to a math contest problem.

I understand that if for both circles, if you assume that a circle's centre, the two points on the circumference that touch a tangent line each, and the intersection of the tangent lines; that these four points form a square, then the distances can be calculated trivially.

Thus, I would like to know how it is known that the points form two squares.

To clarify, this is what I would like to know:

Given a circle and two perpendicular lines both tangent to the circle, how is it known that the tangent points + center + intersection point forms a square?

• I don't understand your question, are you asking, given a circle and two perpendicular lines both tangent to the circle, how the tangent points + center + intersection point forms a square? Mar 29 '17 at 21:39
• @mdave16 Exactly. Mar 29 '17 at 21:42

You are given that the angle where tangents intersect is $90^\circ$. Concentrate on the larger circle, specifically where the upper tangent contacts it. By the definition of a tangent you know that the tangent is at right angles to radius drawn there. So that is another $90^\circ$. The same argument applies at the contact of the lower point. That's $270^\circ$ total. Now, the intersection of the two tangents together with the two contact points and the circle center form a quadrilateral. The interior angles of any quadrilateral add to $360^\circ$. We have accounted for $270^\circ$, leaving $90^\circ$. So we have a rectangle, at least. But the two radii are equal in length. So we have a square.

It is given that the two tangents form a 90 degree angle with each other. it is also known that the angle formed by the tangent to a circle and the radius to that point is 90 degrees as well. Since the figure emerging is a quadrilateral, the sum of the interior angles is 360 degrees, so the last remaining angle formd by the two tangent points and the center must be 90 degrees as well.

The distance from the center to the two tangent points is the radius and is thus equal. So we now have a quadrilateral where all the angles are 90 degrees and two adjacent sides are equal. So one can conclude it must be a square.

Note that this is not a 'proof', more of an intuition or a way to 'see' why it is a square. That is what I understood from your question.