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Given a circle y with center C and radius r, and a fixed point F inside the circle. For any point Q on the circumference, we consider the line CQ (a radius) and the perpendicular bisector of QF. The intersection of these two lines is point P. We need to show that P lies on an ellipse with foci C and F, and determine the length of the major axis of this ellipse.

1. Properties of the Perpendicular Bisector: Since P is on the perpendicular bisector of QF, it follows that PQ=PF.
2. Collinearity on CQ: Since $P$ is on the line CQ, the points C, P, and Q are collinear, and we have PC+PQ=CQ=r
3. Equating Distances: From the collinearity, PQ=r−PC. Combining this with the perpendicular bisector property PQ=PF, we get PF=r−PC.
4. Sum of Distances: Adding PC and PF, we find PC+PF=PC+(r−PC)=r.

This shows that the sum of the distances from $P$ to the foci $C$ and $F$ is constant and equal to r, which is the definition of an ellipse with foci at C and F. The major axis of an ellipse is the constant sum of the distances from any point on the ellipse to the two foci, which in this case is r.

Thus, the length of the major axis of the ellipse is r​.