Three Circles and a Minimum

★ 7/10algebrainequality

Problem

Real numbers x,y,tx, y, t satisfy

x2+y2=1,(x−t)2+y2=5.x^2 + y^2 = 1, \qquad (x-t)^2 + y^2 = 5.

Find the minimum value of x2+(y−2t)2x^2 + (y - 2t)^2.

Answer

Solution

Difficulty7/10
Topicsalgebra, inequality

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