# 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.