引用第20樓kylau於2007-05-15 00:58發表的“”: Centre of circle at (h,0) So eq of circle is (x-h)^2 + y^2 = r^2 .............(*) Intercept with 2x-y-1=0 at (1,1) So sub (1,1) into (*) get one equation For line 2x-y-1=0, slope = 2 So use slope between (0,h) and (1,1) = -1/2 to have the second equation. Solve for h first, then r, then eq of circle is done. .......
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