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Last edited by hbghlyj 2020-4-13 13:13已知圆$C:(x-a)^2+(y-b)^2=r^2(r>0)$与x轴和直线$y=\sqrt3(x+1)$均相切,且$\led
\sqrt3a-b+\sqrt3&\ge0\\\sqrt3a+b-\sqrt3&\le0\\b&\ge0\endled$,则$\frac{ab+r^2}{a^2+b^2}$的最小值为
将$a=\sqrt3b-1$代入，解得$0< b\le \frac{\sqrt3}2$，$\frac{ab+r^2}{a^2+b^2}=\frac{b (-1 + (1+ \sqrt3) b)}{1 - 2 \sqrt3 b + 4 b^2}\ge\frac{1}{2} \left(\sqrt{2}+1\right)$，当$\frac{1}{2} \left(\sqrt{2-\sqrt{3}}+1\right)$时取等 |
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