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At first, let's name all the circles centers as X₁, X₂, etc. So the first thing we need is to join each X₁, X₂, etc. circles centers with both A and O points. … Proof-steps are the same for all the other circles center. …

I have the next exercise: Find the locus of the centers of the circles that pass through "these" two points Here is the image with my solution drawing: Sure, we can see that the locus I have to find … is the mid perpendicular, but here is the point: to prove that this is the mid perpendicular I have to prove at first that all these centers of the circles are on the same straight line. …

To do this, draw a tangent to the two given circles, that they both will share, through the point K. … By a theorem in my book(which I think is a commonly known fact), since the $LJ$ line is tangent to both circles: $\angle LKA = \frac{1}{2} \cup AK = \frac{1}{2} \cup EK \implies \angle KFE = \angle KBA …

In a textbook the following problem example: Two circles touch at point $M$. … How to prove that the line $CD$ is tangent to both of the circles at the point $M$, when it's tangent to one of them? P.s. …

The problem Two circles are inscribed in the corner. … The segment $AB_{1}$ intersects these circles at points $C$ and $C_{1}$. …

The problem: In an acute-angled triangle $ABC$, point $H$ is the orthocenter, point $O$ is the center of the circumscribed circle, point $J$ is the center of the inscribed circle, $\angle BAC = 60°$. …

The problem: Given two circles with different radii, they intersect each other only at one point $M$. The smaller circle is inscribed into the bigger circle. …