I faced the above question in my textbook-
Two circles with centres O and Q intersect at two points, one being A. Their tangents at A meet the circles again at B and C respectively. A point P is constructed such that AOPQ is a parallelogram. Prove that P is the circumcentre of triangle ABC.
The method to solve this question is fairly simple. You apply the theorem that tangents are perpendicular to radius at point of contact, and then that the point P lies on the perpendicular bisectors of two sides of the triangle.
For context here's a formal proof- https://www.toppr.com/ask/question/let-a-be-one-point-of-intersection-of-two-intersecting/
However, when I drew the diagram of this question, often P was in the exterior of triangle ABC or on it. For reference, here's the diagram given along with the question in the textbook :
That being said, what is the condition for P being within the triangle ABC? This is my question. Does this condition relate to the circles, as in, can we provide a condition including the radii of the circles, and the distance between the lens of the intersection? Note that any condition without some of these terms also works.
EDIT:-
From a discussion in the comments, P will lie on BC if the circles were orthogonal. We know that by the Pythagorean theorem, two circles of radii r_1 and r_2 whose centers are a distance d apart are orthogonal if
r_1^2+r_2^2=d^2.
So what you have to essentially find out is-
- Threshold cases is the other two angles are acute.
- How to exactly express this in the required terms.
- If that's not possible, then why?