MathDB
The Minumum Value

Source: 2002 National High School Mathematics League, Exam One, Problem 2

March 12, 2020

Problem Statement

Real numbers x,yx,y satisfy that (x+5)2+(y−12)2=142(x+5)^2+(y-12)^2=14^2, then the minumum value of x2+y2x^2+y^2 is (A)2(B)1(C)3(D)2\text{(A)}2\qquad\text{(B)}1\qquad\text{(C)}\sqrt3\qquad\text{(D)}\sqrt2\qquad
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